Integration is a part of Calculus, which means joining many small parts under certain area and it is a reverse process of differentiation. There are many Integration methods which are use for calculating integration of Functions, they are given as: 1. General methods: By the help of this integration method, we calculate integration of constant function- ∫cdx = cx + C. 2. General power method:
∫x n+1 With the help of this integration method we calculate integration of function which has variables in power. These methods are use for calculating integration of all variables which have certain power. Product method: ∫ f(x).g(x) dx = f(x).∫g(x).dx – g(x).∫f(x).dx. With the help of this method we calculate integration of two functions which are written in multiply form. 4. Integration of 1/x: Integration of 1/x is an exception because it produces ln x as a result. ∫1/x dx = ln x + C. 5. Integration of Special Functions: There are many special functions. a) Exponential function: integration of Exponential Function is -
∫e
∫a ln a b) Trigonometric function: Integration of Trigonometric Functions is - ∫sin x dx = cos x + C, ∫cos x dx = -sin x + C. c) Hyperbolic functions: Integration of Hyperbolic Functions is - ∫sin hx dx = cos hx + C, ∫cos hx dx = sin hx + C. These are some basic methods of integration which are useful for calculating integration of functions. So, this is all about integration methods. |

Integration by substitution plays a very important role whenever you deal with integrations in Calculus. Follow the steps given below to do Integration substitution:

1. Write the given integral first, which will be in the form of,

∫ f(x)dx,

2. Now substitute the given function with t=t(x),

3. Integrate the function as dx=d/dx(x(t))*dt,

4. Just make the substitution as ∫ f(x)dx=∫ f(t)dt,

5. Substitute the value of t in terms of x to get the final answer.

Now, we will see the following examples given below to get the clear idea about substitution integration.

Example 1: Solve the given examples by substitution integration, f(x)= ∫ sin(4x+5)dx?

Solution:

For solving this type of problem you need to have knowledge about integration by substitution with this you must have knowledge of integration of trigonometric Functions. Now, let's see how to solve this type of problem.

Let 4x+5=t

We will differentiate this function to get the value of dx,

=> 4(1)dx+0=dt,

=> dx=dt/4,

Now we will substitute the value of t and dx in the above equation,

Now above equation can be written as,

=> f(x)= ∫ sin(t)dt/4,

=> f(x)=1/4(-cos(t)) +c,

Now we will substitute the value of t to get the desired answer.

=> f(x)=1/4(-cos(4x+5))+c,

This is the required answer for the given problem.

Example 2: Solve the given examples by substitution integration,

=> f(x)= ∫1/ 2x+3?

Solution: The above mention problem seems very difficult because you can't solve this problem by Integration by parts but when you apply integration by substitution it can easily solve.

Let 2x+3=t,

Now, we will differentiate this function to get the value of dx,

=> 2(1) dx+0=dt,

=> dx=dt/2,

We can replace dx as dt/2,

Now we will put the value of t and dt to get the desired result,

=> ∫1/t *dt/2,

=> 1/2 ∫1/t*dt,

We integrate t as ln(t), now we can rewrite the given equation as:

=> 1/2(ln(t)) +c,

Now we will replace t by 2x+3,

=> 1/2*ln(2x+3)+c,

This is the required answer for the given problem.

Now we will see some complicated examples of integration by substitution in which we will also apply other rules of integration with the above described method.

Example 3: Solve the given problem by integration by substitution,

∫ xcos(3x2+5)dx?

Solution:

As in the above problem there are two Functions so we need to apply Product rule, with this we need to apply the integration by substitution.

∫xcos(3x^{2}+5)

We will take x as first function and cos(3x^{2}+5) as a second function. Now you can ask one question that why we have taken x as first function and cos(3x2+5) as second. to solve this problem we will take 1^{st} function to the problem whose differentiation is simple.

=> x* ∫cos (3x^{2}+5)dx ∫ d/dx(x)* ∫cos(3x^{2} +5)dx,

For doing the integration of cos(3x2+5), we need to apply integration by substitution nad on doing so we get:

=> 3x^{2}+5= t,

=> 6x *dx +0=dt,

=> dx= dt/6x,

=> x*sin(t)dt/6x *sin(t)*dt/6,

x gets cancel with x now we have,

=> sint *1/6xsint,

Now, we can put the value of t in the given equation to get the desired answer.

=> (sin(3x^{2}+5)*sin(3x^{2}+5))/6,

Example 4: Solve the given equation by substitution ∫ log tan^{-1}x/(1+x^{2})dx?

Solution: We can easily see that this problem is complicated and we can't get its solution using by part, but if you are having good knowledge of integration of trigonometric function then we can see that d/dx of tan^{-1}x is 1/(1+x^{2}) so we will take tan^{-1}x as t , and when we will find the value of dt so that 1/(1+x^{2}) cancel out.

Let t= tan^{-1}x,

We will differentiate this function to get the value of dt,

=> dt= 1/(1+x^{2})dx,

=> ∫ log t/(1+x^{2})*(1+x^{2}),

Now we will have,

=> ∫ log t+c,

Now as we know log t can be integrated as 1/t,

=> 1/t +c,

we can replace t by tan^{-1}x to get the answer,

=> 1/ tan^{-1}x +c,

This is the required answer.

Example 5: Solve the given equation by substitution

∫ cos(logx)/x dx?

Solution:

In this question we can either take 1/x as t or we can take logx as t. But when we will take log x as t we can see that 1/x gets cancel with x and our problem is getting simpler so we will take log x as t.

Let logx=t,

=> 1/x*dx=dt,

=> dx=x*dt,

On putting the value dx and t in the above equation we will get:

=> ∫cos(t)/x*x*dt,

x gets cancel with x and we will get,

=> ∫cost*dt,

=> -sint +c,

Now we can put the value of t to get the desired answer.

-sin(log(x))+c,

This is the required answer.

Example 6: Solve the given equation by substitution

f(x)= 1/x^{2}+3x+4?

Solution:

For solving this type of problem we have to apply substitution method.

Let t=x^{2}+3x+4,

=> dt=2x+3+0,

=> dt=(2x+3)dx,

Now we can rewrite the equation as,

f(x)= ∫1/t (dt)*(2x+3),

f(x)= logt*(2x+3) +c,

We will put the value of t to get the desired answer,

f(x)=log(x^{2}+3x+4)*(2x+3),

This is the required answer.

Example 7: Solve the given equation by the method of integration by substitution method

(1+cosx)^{7}*sinx(dx)?

Solution: For solving this type of problem we have to apply substitution method. This type of problem can easily be solved if we know which function have to assign with t.

We will take bigger function as t because by assuming it equal to t we can minimize the given function. So we will take

1+cosx= t,

We will differentiate this function to get the value of dx,

=> 0+sinx(dx)= dt,

=> dx=dt/sinx,

=> dt=dx(sinx),

Now we will put the given values in the above equation we will get:

=> t^{7}*sinx*1/sinx *dt,

sinx gets cancel with sinx we will get:

=> ∫t^{7}dt,

=> t^{8}/8 +c,

We will put the value of t in the above equation we will get:

=> (1+cosx)^{8}/8 +c.

This is the required answer.

In Calculus the study of Integration by parts is very important. In mathematical analysis or in calculus, integration by parts is a rule which replaces the integral of products of Functions into the other ideally simpler integrals. This rule can be derived from the Product rule of differentiation. By using the formula of integration by parts we can find the integral of two different Functions. The formula is u dv = uv - ? v du, which is expressed as below and where u and v are differentiable functions. Suppose that if u = f(x), v = g(x), then the differentials of this function is du = f '(x) dx and dv = g'(x) dx, and from this the integration by parts states that:

∫u dv = uv-∫v du

Or,

∫f(x) g^{’}(x) dx = f(x) g(x) - ∫f^{’}(x) g(x) dx

For example we are taking a function

∫ xcos(3x^{2}+5)dx

As we are seeing in the above problem we apply integration by parts,

∫xcos(3x^{2}+5)

Now, we will take x as first function and cos(3x^{2}+5) as a second function. Now you can ask one question that why we have taken x as first function and cos(3x2+5) as second . Answer of this question is that we will take 1^{st} function to that function whose differentiation is simple.

x* ∫cos (3x^{2}+5)dx ∫ d/dx(x)* ∫cos(3x^{2} +5)dx

Now for doing the integration of cos (3x^{2}+5), we need to apply Integration by substitution

3x^{2}+5= t,

6x *dx +0=dt,

dx= dt/6x,

x*sin(t)dt/6x *sin(t)*dt/6,

x gets cancel with x now we have:

sint *1/6xsint,

Now we can put the value of t in the given equation to get the desired answer:

(sin(3x^{2}+5)*sin(3x^{2}+5))/6,

From the above formula u is the first function and v is the second function. These are of any of the types like algebraic, trigonometric, inverse trigonometric, logarithmic and exponential .In this method first function u is differentiated and the second function v is to be integrated. The easiest way to remember to select the function as which one is first function or which one is the second function we follow a rule that is explained below,

Suppose that we represents algebraic function with the starting character A, trigonometric as T, inverse trigonometric as I, logarithmic as L and exponential as E respectively. When we combine these letters it become "LIATE " so from this which letter comes first we can solve first that is the first function is to be selected which comes first in the order of the letters "LIATE " .This rule of thumb proposed by Herbert Kasube of Bradley University and all functions are explained below

L: Logarithmic functions are ln x, log_{b} x, etc.

I: Inverse Trigonometric Functions are arctan x, arcsec x, etc.

A: Algebraic functions are x^{3}, 2x^{23}, etc.

T: Trigonometric functions are sin x, tan x, etc.

E: Exponential functions are e^{x}, 29^{x}, etc.

The function in dv comes last in the list the functions lower on the list have easier anti Derivatives than the functions above them. This rule is also written as "DETAIL" where D stands for dv. Lets an example to understand the LIATE rule, we consider the integral

∫-xcosx dx

We follow the LIATE rule, u = -x and dv = cos x dx , hence du = -dx and v = sin x , which makes the integral

-xsinx - ∫-sin x dx which is equal to

-xsinx - cosx +c

In other words, we try to choose u and dv such that du is simpler than u and then dv is easy to integrate. If we use u instead of cos x and x as dv, we would have the integral as follows

-x^{2}/2 cosx + ∫-x^{2}/2 sinx dx

The recursive application of the integration by parts formula would result in an infinite Recursion whereas rule of thumb, there is exceptions to the LIATE rul. We consider the rules in the "ILATE" order instead. In some cases, polynomials are needed to be split in non-trivial ways.

If we suppose that f(x) and g(x) are two continuously differentiable functions then the product rule is

∫ [f(x) g(x)]^{’}= f(x) g^{’}(x) + f^{’}(x) g(x)

Now, integrating both sides the above function

f(x) g(x) = ∫f(x) g’(x) dx + f^{’}(x) g(x) dx

And

∫ f(x) g^{’}(x) dx = f(x) g(x) - ∫f^{’}(x) g(x) dx

So, from the above we can derive the integration by parts rule and given an interval m and n,

_{m}∫^{n} f(x) g^{’}(x) dx = [f(x) g(x)]_{m}^{n} -_{m}∫^{n} f^{’}(x) g(x) dx

[f(x) g(x)]_{m}^{n}= f(n) g(n) - f(m) g(m)

This shown to be true by using the product rule for derivatives so that

f(n) g(n) - f(m) g(m) = _{m}∫^{n} d/dx f(x) g(x) dx

=_{m}∫^{n} f^{’}(x) g(x) dx + _{m}∫^{n} d/dx f(x) g^{’}(x) dx

So that the above the rule is often stated using indefinite integrals in the form of

∫ f(x) g^{’}(x) dx = f(x) g(x) - ∫f^{’}(x) g(x) dx

Or in the other words if u = f(x), v = g(x) and the differentials du = f ′(x) dx and dv = g′(x) dx,

∫u dv = uv-∫v du

From the above formula it means that the area under the graph of a function u (v) is the same as area of the Rectangle above of the graph with u v minus.

In order to apply the rule the original integral contains the derivative of g and the anti derivative of g must be found and then the integral ∫g f ′ dx can be evaluated. For discrete analogue for sequences one formula can used and is called summation by parts. The original expression of f(x) and g(x) is expressed as

_{m}∫^{n} f(x) g(x) dx = f(x) ∫g(x)dx]_{m}^{n} -_{m}∫^{n} (∫g(x dx) f^{’}(x) dx

This formula is applicable when f(x) is continuously differentiable and also g(x) is continuous. These are known as the Riemann–Stieltjes integral and Lebesgue–Stieltjes integral. And more complicated forms of the above rule are also valid for calculation and is given below,

∫xy dz= xyz - ∫xz dy - ∫yz dx.

This formula is used for performing the integration on the product. The integration of the product can be easily integrated if one of the products is unity. If the product of two functions is different then we can simply use the concept of integration by parts.

Integration by parts not purely a mechanical process for solving integrals but integration by parts is a heuristic given a single function to separate it into a product of two functions f(x)g(x) such that the integration by parts formula is easier to calculate than the original one and formula is shown in below

∫fg dx = f∫g dx - ∫ (f^{’}∫g dx ) dx

From the above right-hand side, the function f is differentiated and function g is integrated or it is useful to choose f as a function that simplifies when differentiated, and to choose g as function that integrated.

Now, one of the topic which is related to the integration by parts that is calculus of variations. Calculus of variations is a field of mathematics which deals with maximizing or minimizing of the function as ordinary calculus which deals with maximizing and minimizing of ordinary functions. In calculus of variations a functional is usually a mapping from a Set of functions. These functions are formed by definite integrals involving unknown functions and their derivatives. In external functions that make the functional a maximum or minimum value those where the rate of change of the functional is zero.

The simplest example of this type of problem is to find the curve of shortest length connecting two points. The solution is straight line between the points if there are no constraints. On other hand if the solution is less obvious and possibly many solution then curve is constrained to lie on a surface in space and these solutions are called geodesics. This problem is posed by Fermat's principle that is light follows a path of shortest length connecting two points, where the length depends upon the material of the medium in which the light travels. Many important problems come up when function has several variables. Solutions of that kind of problems we use the Laplace equation with the Dirichlet principle. Now I am taking an example that shows the Integration by Parts

∫(xcosx)dx

Let u=x and dv = cosx dx

Then du = dx and u = sinx therefore the above equation become by using integration by parts formula

∫(xcosx)dx=x(sinx)- ∫sinx dx

Now we solve the equation

x(sinx) + cosx +c , where c is the constant which belong to the integration. I am telling you some important points which is useful to solve the problems which are related to integration by parts.

When we solving the any function by using integration by parts we choose a function such that the function u reduces which depends upon differentiation while other function dv remains integrable on integration. For example we are taking the function is x^{3}cosx then we will choose u as x^{3} and then applying integration by parts till it becomes1. The other function dv, which is cosx dx becomes sinx which is equally integral. For example

_{0}∫1-x^{2} exdx

Firstly we find out the Point that has a Definite Integral. So the answer will be a number not a function of x and the integral of e^{x} leads to be same function. It is not a big deal that we do one operation or the other. Now, we concentrate on the other function x^{2}. Therefore, if we integrate the function then we will increase the power. From this we should differentiate x^{3} and integrate e^{x}. And the above equation become

u=-x^{2}

dv = e^{x}dx

Now we integration and differentiation the function and we get

du =-2xdx

v=e^{x}

And integration by parts formula gives us

_{0}∫^{1}-x^{2}e^{x}dx = [-x^{2}e^{x}]_{0}^{1}- _{0}∫^{1} 2x^{2} e^{x}dx

So, that the new integral _{0}∫^{1} x e^{x}dx is not easily obtainable. In this case we will use integration by parts again. This is same discussion as before that is

u=-x

dv = e^{x}dx

This implies

du = dx

v = e^{x}

The integration by parts formula gives us

_{0}∫^{1}-xe^{x}dx = [-xe^{x}]_{0}^{1}- _{0}∫^{1} -e^{x}dx

_{0}∫^{1} -e^{x}dx=[-e^{x}]_{0}^{1},we get

_{0}∫^{1}-xe^{x}dx=[-xe^{x}]_{0}^{1}-[-e^{x}]_{0}^{1}

Finally it is easy calculate

_{0}∫^{1}-x^{2}e^{x}dx = [-x^{2}e^{x}]_{0}^{1}-2[-xe^{x}]_{0}^{1}+2[-e^{x}]_{0}^{1}

After solving the above equation we get the final answer that is

_{0}∫^{1}-x^{2}e^{x}dx=e+2

From this example conclusion is that we remember that many times the integration by parts will not be enough to give us the answer after one shot. We should need to do some extra work that is integration by parts or use other techniques like Laplace.

The other method of calculus integration by parts is interchange of the order of integration, from the above formulation we includes the technique of interchange of the order of integration. Let us consider iterated integral function,

_{a}∫^{b} dx _{a}∫^{x} dy k(y)

In the order written the above formula, the width dx is integrated first over the y-direction of width dx in the x direction is integrated with respect to the y across the y direction, when the function k(y) is not easily integrated. The integral of the given function can be reduced to a single integration by reversing the order of integration. The strip of width dy is first integrated from the line x = y to the limit x = b to accomplish this interchange of variables and the result is integrated from y = a to y = b and written as

_{a}∫^{b} dx _{a}∫^{x} dy k(y)=_{a}∫^{b} dy k(y) _{a}∫^{x} dx=_{a}∫^{b}dy(b-y)k(y)

The result can be seen in the above formula for integration by parts and repeated below,

_{a}∫^{b} f(x) g^{’}(x) dx = [f(x) g(x)]_{c}^{b} -_{a}∫^{b} f^{’}(x) g(x) dx

Substitute

f(x)=_{a}∫^{x} dy k(y) and g^{’}(x)=1,Which gives us the final result.

The recursive integration by parts is shown in below formula

∫uv=uv_{1}+u’v_{2}+u’’v_{3}+………+(-1)^{n} u^{(n)}u_{n+1}

In the above u^{’} shows the first derivative of u and u^{’’} shows the second derivative. Furthermore u^{(n)} represent the its nth derivative with respect to the independent variable. We go to another approved notation in the calculus that is

u_{n+1}(x)= ∫∫ ….. ∫v (dx)^{n+1}

*n* + 1 means there are n +1integrals.

In the above equation Integrand (*uv*) differs from the previous equation. dv factor has been written as *v* for convenience. This is because it can be evaluated by differentiating the first term and integrating the second with sign is changed every time and starting out with uv_{1}. This is very useful factor specially in cases when *u*^{(}^{k}^{+1)} becomes zero for some *k* + 1.Therefore once the *u*^{(}^{k}^{)} term has been reached the integral evaluation can stop.

When the recursive definition is correct, then it is often tedious to remember and implement. The much easier representation of this process is taught and is dubbed either the stand and deliver method rapid repeated integration or the tic-tac-toe method. This method is very useful when one of the two functions in the product is a polynomial and after differentiating it several times one obtains zero. For example

∫x^{2}cosx dx

Let u = x^{2}. Now begin this function and list in a column all the derivatives until zero is reached. And secondly begin with the function v with cos(x) and list each and every integral of v until the size of the column is the same as that of u. The result is shown as follows

Derivatives of u (Column A) Integrals of v (Column B)

X^{2} cosx

2x sinx

2 -cosx

From the above table, the 1st entry of column A with the 2nd entry of column B, the 2nd entry of column A with the 3rd entry of column B, etc. With the alternating signs beginning with the positive sign. We can calculate the function until pairing leads to sums of zeros.

Now the example of whole discussion, we integrating xe^{x} we need to apply integration by parts. From the above f(x) and g(x) be the two function and we marked f(x) as 1st function and g(x) as second function, Now according to integration by parts

∫f(x) g^{’}(x) dx = f(x) g(x) - ∫f^{’}(x) g(x) dx

Now, one question arises that which function we have to take as first function and which function as second

It’s very simple here we use LIATE. We will take 1st function to that function whose differentiation is easy, we take this just for simplifying the expression which have double integral.

Now we will move to problem

∫ x*e^{X}

In this problem we can take any function as 1st function, because we know differentiation of both the function is very easy, as we know that differentiation of x is 1 and differentiation of e^{X} is e^{X}.

But for the above problem we will take x as first function and e^{X} as second function, we are taking this just because if we take e^{X} as first function then we will face problem in double integral section. As we know that differentiation of e^{X} is e^{X} and integration of x will be x^{2}/2, so we will have two functions again and we have to apply integration by parts again.

∫ x*e^{X}=x*∫ e^{X} -∫d/dx(x)*∫ e^{X}

=x*e^{X}-∫ *e^{X} +c1

=x*e^{X}-∫ e^{X} +c1

=x*e^{X}-e^{X}+c1 +c2

=e^{X}(x-1) +c1 +c2

Where c1 and c2 are the constant of integration

This is the required answer. We have two constant c1 and c2 here just because we have integrated the function two time, we can solve this type of problem we just need have the knowledge of integration by parts.

The next example is we integrating 2xe^{x}. From the above f(x) and g(x) be the two function and we marked f(x) as 1st function and g(x) as second function, Now according to integration by parts

∫f(x) g^{’}(x) dx = f(x) g(x) - ∫f^{’}(x) g(x) dx

Now, one question arises that which function we have to take as first function and which function as second

It’s very simple here we use LIATE. We will take 1st function to that function whose differentiation is easy, we take this just for simplifying the expression which have double integral.

Now, we will move to problem,

∫ 2x*e^{X}

In this problem we can take any function as 1st function, because we know differentiation of both the function is very easy, as we know that differentiation of x is 1 and differentiation of e^{X} is e^{X}.

But for the above problem we will take x as first function and e^{X} as second function, we are taking this just because if we take e^{X} as first function then we will face problem in double integral section. As we know that differentiation of e^{X} is e^{X} and integration of x will be x^{2}/2, so we will have two functions again and we have to apply integration by parts again.

∫ 2x*e^{X}=2x*∫ e^{X} -∫d/dx(2x)*∫ e^{X}

=2x*e^{X}-∫ 2*e^{X} +c1

=2x*e^{X}-∫ e^{X} +c1

=2x*e^{X}-2e^{X}+c1 +c2

=2e^{X}(x-1) +c1 +c2

Where c1 and c2 are the constant of integration

This is the required answer. We have two constant c1 and c2 here just because we have integrated the function two time, we can solve this type of problem we just need have the knowledge of integration by parts.

Now, I am tell you some other integration by parts methods which are named as Lebesgue–Stieltjes integration generalizes Riemann–Stieltjes and Lebesgue integration. There are many advantages of these methods in the measure of theoretic framework. In ordinary Lebesgue integral with respect to a measure known as the Lebesgue–Stieltjes measure that may be associated to any function. The Lebesgue–Stieltjes measure is a regular measure and conversely every regular measure on the real line.

Lebesgue–Stieltjes integrals, named for Henri Leon Lebesgue and Thomas Joannes Stieltjes, are also known as Radon integrals. They are used in Probability and stochastic processes and also in certain branches of analysis including potential theory. The Lebesgue–Stieltjes integral is defined as

_{a}∫^{b}f(x) dg(x)

, when function f: [a, b] → R is measurable and bounded and g is [a, b] → R is of variation in [a, b] and right-continuous. When f is non-negative and g is monotone and right-continuous. We assume that f is non-negative and g is monotone non-decreasing and right-continuous. As k ((s, t]) = g (t) − g(s) and k (a) = 0 and the construction works for g left-continuous, k([s, t)) = g(t) − g(s) and k(b) = 0.

u_{i}(E)=inf∑u_{g}(I_{i})|E СUI_{i}

From the above Lebesgue–Stieltjes integral

_{a}∫^{b}f(x) dg(x)

In other case if g is non-increasing then Lebesgue–Stieltjes integral is

_{a}∫^{b}f(x) dg(x) =-_{a}∫^{b}f(x) d (-g)(x)

This is defined by the preceding construction.

Consider if g is of bounded variation and f is bounded then we can write

g(x) = g1(x) − g2(x)

Where g1(x) is the total variation of g in the interval [a,x], and g2(x) = g1(x) − g(x). g1 and g2 both are monotone non-decreasing. Now the Lebesgue–Stieltjes integral with respect to g is defined as

_{a}∫^{b}f(x) dg(x) = _{a}∫^{b}f(x) d (g1)(x)-_{a}∫^{b}f(x) d (g2)(x),where the two integrals are defined by the preceding construction.

In this article we studied the calculus integration by parts or integration by parts calculus.

The Integration Partial Fractions is an important function in the integration. But first of all the definition of integration, the integration is just reverse function of differentiation or integral is an anti derivative function whose derivative is any function. Both Functions i.e. integration and differentiation are important factor in mathematics. These both are the two main operations that we perform in Calculus and it is known as indefinite integral. Mathematical expression of indefinite integral is given by:

_{a}∫

^{b}f(s) ds

These are the basic tools of calculus with wide applications in science and engineering field. Any line of integral is defined for Functions with two or three variables and having interval (x, y) is changed by two points on the plane where they meet and create a curve. These factors play an important role to develop the many physics law which are electrodynamics. Integration Partial Fractions is one of the major part of integration.

By using partial fraction we find the integration of a Rational Function in integral calculus. In integration by partial fractions we can write any rational number as a finite number of algebraic fractions and also the sum of given parameter of function and these parameters are called polynomial function. The degree of polynomial function in some of positive Integer the power of denominator a polynomial function is 1 or 2. In other word rational function of a Complex Number all denominators will have a polynomial of degree 1.

If the denominator has 1 degree polynomial then we can say that the numerator is constant. On other hand when the numerator has 1 degree then the denominator is a 2 degree polynomial. If the integral function is in the form of an algebraic fraction and the integral cannot be calculated by simple methods then we can simplify the function by partial fractions before integration. To do this first we split the given algebraic fraction into its partial fractions and the result is considered as addition or subtraction. It is necessary for a fractional function the numerator must have one degree less than the denominator. To compute integrals of the form that is,

∫M(x)/N(x) dx

When we decompose the ∫M(x)/N(x) dx this is called partial fraction. We can write these function as rational functions as f(x) = M(x)/N(x). In partial fraction expansion first we make sure that deg M(x) < deg N(x) by using Long Division. There are four possible cases which increasing generality and difficulty and these are,

The product of distinct N(x) is linear factors.

The product of distinct N(x) is linear factors, some of which are repeated.

The product of distinct N(x) is irreducible quadratic factors, some of which may be repeated.

N(x) is has repeated irreducible quadratic factors, there is a possibility that some linear factors which may repeat.

Now the Integration Partial Fractions theorem; suppose that a, b

_{1 }and b

_{2}are Polynomials that are relatively prime functions and having different factor. The polynomials are, x

_{1}and y

_{2}such that:

a/ b

_{1}b

_{2}= x

_{1 }/ b

_{1}+ y

_{2 }/ b

_{2}

Proof of the above theorem a

_{1 }and b

_{2}are prime, using the long division method, we can find polynomials p

_{1}and p

_{2}such that

1= p

_{1}b

_{1 }+ p

_{2}b

_{2}

Dividing both sides by n

_{1,}n

_{2}and multiplying by m we get:

a/ b

_{1}b

_{2}= x

_{1 }/ b

_{1}+ y

_{2 }/ b

_{2}

Hence the Integration Partial Fractions theorem proved.

While doing factor of M(x), we can write the irreducible quadratic factors in the form ax

^{2}+ bx + c with b

^{2}− 4ac < 0). For example if M(x) = 3x

^{2}-3x-6 is Quadratic Equation and we can factor the M(x) as:

M(x) = 3x

^{2}-3x-6 =3(x

^{2}-x-2) =3(x

^{2}+x-2x-2),

=3(x(x+1)-2(x+1)) =3(x-2) (x+1),

= 3(x-2) (x+1) =0.

We can factor the quadratic as b

^{2}− 4ac < 0, if the function is not like that then we will factor that quadratic equation again. This is shown in above example. Now I will show you how to write M(x) / N(x) as a sum of terms of the form.

A/(ax +b)

^{k}and Ax + B/(ax + bx +c)

^{k}

If m(x) is a product of linear factors with no repeating factor means that M(x) = (a

_{1}x + b

_{1}) (a

_{2}x + b

_{2})...(a

_{n}x + b

_{n}) here no factor is repeated and no factor is a multiple of another. For each linear term we write it in form that is A/(ax +b) and the whole equation is written as:

M(x)/N(x) =A

_{1}/ (a

_{1}x + b

_{1}) + A

_{2}/(a

_{2}x + b

_{2}) +….+ A

_{n}/ (a

_{n}x + b

_{n})

If M(x) is a product of linear factors and some of which are repeated then (ax + b) appears in the Factorization of M(x) k times then we can using the complicated method instead of pieces,

A

_{1}/ (a x + b) + A

_{2}/ (ax + b)

^{2}+….+ A

_{n}/ (a

_{n}x + b

_{n})

^{n}

For example consider the following addition of algebraic fractions,

1/(x+1) + 3/(x + 2) = [(x+2) + 3(x+1)]/ (x + 2) (x+1),

= (4x + 5)/x

^{2}+3x +2.

From the above we start with the expression (4x + 5)/x

^{2}+3x +2 and then find the fractions whose sum gives the result. Then we find the two fractions that is 1/(x+1) and 3/(x + 2), are called the partial fractions of (4x + 5)/x

^{2}+3x +2.

We split the fractions like above functions because of the reason, that makes integrals much easier to do and other method is used that is the Laplace Transform, which is discussed later. So that the above function is,

∫ (4x + 5)/x

^{2}+3x +2 dx =∫1/(x+1) + ∫3/(x + 2),

Now it easy to integrate,

∫ (4x + 5)/x

^{2}+3x +2 dx= ln(x +1) + 3 ln(x+ 2) + k,

The now the 1-degree polynomial in the denominator is

We substitute u = ax + b, du = a dx , this reduces the integral,

∫1/ [ax +b] dx,

∫1/u du/a = 1/a∫du/u=1/a ln [u] + k = 1/a ln [ax +b] + c.

So this is all about the partial fractions integration.

Hello friends, today we are going to study about two most interesting topics of mathematics, trigonometric integrals and trigonometric substitution. Here I will explain the best ways to understand these topics. Firstly, we will start with trigonometric integrals: The trigonometric integrals can be defined as the member of integrals which have Trigonometric Functions in it. There are lots of Trigonometric Functions which are given below:

ʃ cos x dx = sin x + C

ʃ sin x dx = - cos x + C

ʃ sec^{2 }x dx = tan x + C

ʃ cosec^{2} x dx = - cot x + C

ʃ sec x tan x dx = sec x + C

ʃ cosec x cot x dx = - cosec x + C

ʃ tan x dx = ln | sec x | + C

ʃ cot x dx = ln | sin x | + C

ʃ sec x dx = ln | sec x + tan x | + C

ʃ cosec x dx = ln | cosec x – cot x | + C

Let's take some examples of trigonometric integrals:

Example 1: Integrate ʃ sin 2x dx?

Solution:

To integrate the above problem we will use u-substitution method,

Let u = 2x,

so that du = 2dx,

or ( ½ ) du = dx,

Substitute it into the original problem and replace all 'x', now we will get:

ʃ sin 3x dx = ʃ sin u ( ½ ) du,

= ( ½ ) ʃ sin u du,

= ( ½ ) ( - cos u ) + C,

= - ½ cos 3x + C.

Example 2: Integrate ʃ tan 7x dx?

Solution:

To integrate the above problem we will use u-substitution method,

Let u = 7x,

So that du = 7dx,

Or ( 1/7 ) du = dx,

Substitute it into the original problem and replace all 'x', now we will get:

ʃ tan 7x dx = ʃ tan u ( 1/7 ) du,

= ( 1/7 ) ʃ tan u du,

= 1/7 ln | sec u | + C,

= 1/7 ln | sec 7x | + C.

Example 3: Integrate ʃ 3 sec 2x tan 2x dx?

Solution:

To integrate the above problem we will use u-substitution method

Let u = 2x,

so that du = 2dx,

or ( ½ ) du = dx,

Substitute it into the original problem and replace all 'x', now we will get,

ʃ 3 sec 2x tan 2x dx = 3 ʃ sec u tan u ( ½ ) du,

= ( 5/2 ) ʃ sec u tan u du,

= ( 5/2 ) sec u + C,

= ( 5/2 ) sec 2x + C.

Example 4: Integrate ʃ ( sin x + cos x )^{2} dx?

Solution:

Start with squaring the above function, we get:

ʃ ( sin x + cos x )^{2} dx = ʃ ( sin^{2} x +2 sin x cos x + cos^{2} x ) dx,

= ʃ ( ( sin^{2} x + cos^{2} x ) + 2 sin x cos x ) ) dx,

= ʃ ( 1 + 2 sin x cos x ) dx,

= ʃ 1 dx + ʃ 2 sin x cos x dx,

= x + 2 ʃ sin x cos x dx,

Now we will use u-substitution method,

Let u = sin x,

So that du = cos x dx,

Substitute it into the original problem and replace all 'x', now we will get:

x + 2 ʃ sin x cos x dx = x + 2 ʃ u dx,

= x + 2 u^{2}/2 + C,

= x + u^{2} + C,

= x + sin^{2 }x+ C.

Example 5: Integrate ʃ ( 3 + tan x )^{2} dx?

Solution:

Now start with squaring the above function, we get:

ʃ ( 3 + tan x )^{2} dx = ʃ ( 9 + 9 tan x + tan^{2} x ) dx,

= 9 ʃ 1 dx + 9 ʃ tan x dx + ʃ tan^{2} x dx,

= 9x + 9 ʃ tan x dx + ʃ tan^{2} x dx,

= 9x + 9 ln | sec x | + ʃ ( sec^{2} x – 1 ) dx,

= 9x + 9 ln | sec x | + ʃ sec^{2} x dx – ʃ 1 dx,

= 9x + 9 ln | sec x | + ʃ sec^{2} x dx – x,

= 8x + 9 ln | sec x | + ʃ sec^{2} x dx,

= 8x + 9 ln | sec x | + tan x + C.

Now let's move on next topic trigonometric substitutions: The trigonometric substitution can be defined as the substitution of Trigonometric Functions for other expressions. We may use Trigonometric Identities to solve integrals containing radical functions.

If the integral have a^{2} – x^{2}, let,

x = b sin ɵ,

And use this identity,

1 – sin^{2} ɵ = cos^{2} ɵ,

if the integral have a^{2 }+ x^{2}, let,

x = a tan ɵ,

and use this identity,

1 + tan^{2} ɵ = sec^{2} ɵ.

If the integral have x^{2} – a^{2}, let

x = a sec ɵ,

and use this identity,

sec^{2} ɵ – 1 = tan^{2} ɵ.

Let's take some examples of trigonometric substitution Integration:

Example 1: Evaluate the integral ʃ dx / ( x^{2} √ x^{2} - 4 )?

Solution:

The radical √ x^{2} – 4 in this integral suggests a triangle with hypotenuse of length x and base of length 2:

For this triangle, sec ɵ = x/2, we will try the substitution x = 2 sec ɵ. Then ɵ = sec^{−}^{1} ( x/2 ), where we specify0 < ɵ < pi/2 or pi < ɵ < 3pi/2. Note that dx = 2 sec ɵ tan ɵ dɵ and that √ x^{2} – 4 = 2 tan ɵ.

Then,

ʃ dx / ( x^{2} √ x^{2} - 4 ) = ʃ ( 2 sec ɵ tan ɵ )/( ( 2 sec ɵ )^{2} ( 2tan ɵ ) ) dɵ = ʃ ¼ cos ɵ dɵ = ¼ sin ɵ + C

But we know that sin ɵ = ( √ x^{2} – 4 ) / ( x ),

So,

ʃ dx / ( x^{2} √ x^{2} - 4 ) = ( √ x^{2} - 4 ) / 4x + C.

Example 2: Evaluate the integral ʃ ( 1 / ( √ 9 - x^{2} ) )dx?

Solution:

To evaluate the integral ʃ ( 1 / ( √ 9 - x^{2} ) )dx, we make the substitution

x = 3 sin ( u ) , −π/2 < u < π/2,

dx = 3 cos ( u ) du.

Note that we didn't include both u = −π/2 and u = π/2 since the function being integrated is not defined at either x = −3 or x = 3. Then,

ʃ ( 1 / ( √ 9 - x^{2} ) )dx = ʃ ( ( 3 cos ( u ) ) / ( √9 − 9 sin^{2} ( u ) ) ) du,

= ʃ ( ( 3 cos ( u ) ) / ( √9 − 9 sin^{2} ( u ) ) ) du,

= ʃ ( ( 3 cos ( u ) ) / ( 3√1 − sin^{2} ( u ) ) ) du,

= ʃ ( ( cos ( u ) ) / ( √cos^{2} ( u ) ) ) du,

= ʃ ( cos u ) / ( cos u ) du,

= ʃ du,

= u + c,

Now x = 3 sin u implies that u = sin^{−1 }( x/3 ), so we have,

ʃ ( 1 / ( √ 9 - x^{2} ) )dx = sin^{−1 }( x/3 ) + c.

Example 2: Evaluate the integral ʃ ( 1 / ( √ 16 - x^{2} ) )dx?

Solution:

To evaluate the integral ʃ ( 1 / ( √ 16 - x^{2} ) )dx, we make the substitution

x = 4 sin ( u ) , −π/2 < u < π/2,

dx = 4 cos ( u ) du.

Note that we didn't include both u = −π/2 and u = π/2 since the function being integrated is not defined at either x = −4 or x = 4. Then,

ʃ ( 1 / ( √ 16 - x^{2} ) )dx = ʃ ( ( 4 cos ( u ) ) / ( √16 − 16 sin^{2} ( u ) ) ) du,

= ʃ ( ( 4 cos ( u ) ) / ( √16 − 16 sin^{2} ( u ) ) ) du,

= ʃ ( ( 4 cos ( u ) ) / ( 4√1 − sin^{2} ( u ) ) ) du,

= ʃ ( ( cos ( u ) ) / ( √cos^{2} ( u ) ) ) du,

= ʃ ( cos u ) / ( cos u ) du,

= ʃ du,

= u + c.

Now x = 4 sin u implies that u = sin^{−1 }( x/4 ), so we have:

ʃ ( 1 / ( √ 16 - x^{2} ) )dx = sin^{−1 }( x/4 ) + c.