Methods of integration: If we know an object’s instantaneous velocity at each given time, a logical issue arises: can we calculate the object’s location at any given time? There are a number of such practical and theoretical examples or scenarios in which the integration process is involved.

The expansion of integral calculus is the result of attempting to solve the problem of finding a function whenever its derivative is given and the problem of finding the area enclosed by the graph of a function under certain conditions. These two problems result in two types of integrals, such as indefinite and definite integrals, which together make up the Integral Calculus.

In this article, we are going to learn some standard formulae and different methods to integrate the functions.

What are Different Methods of Integration?

Below we have provided different methods of integration:

1. Integration by Substitution
2. Integration by Parts
3. Integration Using Trigonometric Identities
4. Integration of Some particular function
5. Integration by Partial Fraction 

Integration by Substitution

The method of evaluating an integral by reducing it to standard form by a proper substitution is called integration by substitution.

We use the substitution method for the integrals of the following types:

Type I : Integral of the form, $\int f(\varphi (x)){\varphi }^{\prime }(x)dx$,

• Step 1: Substitute $\varphi (x)=t\Rightarrow {\varphi }^{\prime }(x)dx=dt$
• Step 2: This substitution reduces the above integral to $\int f(t)d\mathrm{t}$.
• Step 3: After evaluating $\int f(t)d\mathrm{t}$, substitute back the value of $t$ in terms of $x$.

Type II : Integral of the form, $\int f(ax+b)dx$

1. If $\int f(x)dx=\varphi (x)$, then $\int f(ax+b)dx=\frac{1}{a}\varphi (ax+b)$
We can prove the above result as follows:

• Step 1: Let $I=\int f(ax+b)dx$
• Step 2: Substitute $ax+b=t\Rightarrow dx=\frac{1}{a}dt$.
• Step 3: $I=\int f(ax+b)dx=\frac{1}{a}\int f(t)dt$
• Step 4: $I=\frac{1}{a}\int f(t)dt=\frac{1}{a}\varphi (t)[$ Since $\int f(x)dx=\varphi (x)]$.
• Step 5: Substitute back the value of $t$, So, we have
  $I=\int f(ax+b)dx=\frac{1}{a}\varphi (ax+b)$

2. $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C,n\ne -1\text{. }$

• Step 1: Let $I=\int (ax+b{)}^{n}dx$
• Step 2: Substitute $ax+b=t\Rightarrow adx=dt$
• Step 3: $\int (ax+b{)}^{n}dx=\frac{1}{a}\int t^{n}dt=\frac{t^{n+1}}{a(n+1)}+C,n\ne -1$.
• Step 4: Substituting back the value of $t$, we have $\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C,n\ne -1$.

In short, we can say that the formula for integrals of the form, $\int f(ax+b)dx$ is same as the formula for $\int f(x)d$. We divide the formula by coefficient of $x$ or by derivative of $(ax+b)$ with respect to $x$, i.e. $a$.

Integrals	Formula
$\int \frac{1}{ax+b}dx$	$\frac{1}{a}\log |ax+b|+C$
$\int \sin (ax+b)dx$	$-\frac{1}{a}\cos (ax+b)+C$
$\int e^{ax+b}dx$	$\frac{1}{a}{e}^{ax+b}+C$
$\int a^{bx+c}dx$	$\frac{1}{b}\cdot \frac{a^{bx+c}}{\log a}+C,a>0$ and $a\ne 1$
$\int \sin (ax+b)dx$	$-\frac{1}{a}\cos (ax+b)+C$
$\int \cos (ax+b)dx$	$\frac{1}{a}\sin (ax+b)+C$
$\int {\sec }^2(ax+b)dx$	$\frac{1}{a}\tan (ax+b)+C$
$\int {\mathrm{cosec}}^2(ax+b)dx$	$-\frac{1}{a}\cot (ax+b)+C$
$\int \sec (ax+b)\tan (ax+b)dx$	$\frac{1}{a}\sec (ax+b)+C$
$\int \mathrm{cosec}(ax+b)\cot (ax+b)dx$	$-\frac{1}{a}\mathrm{cosec}(ax+b)+C$
$\int \tan (ax+b)dx$	$-\frac{1}{a}\log |\cos (ax+b)|+C$
$\int \cot (ax+b)dx$	$\frac{1}{a}\log |\sin (ax+b)|+C$
$\int \sec (ax+b)dx$	$\frac{1}{a}\log |\sec (ax+b)+\tan (ax+b)|+C$
$\int \mathrm{cosec}(ax+b)dx$	$\frac{1}{a}\log |\mathrm{cosec}(ax+b)-\cot (ax+b)|+C$

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Type III: $\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C$

• Step 1: Let $I=\int \frac{f^{\prime }(x)}{f(x)}dx$.
• Step 2: Substitute $f(x)=t\Rightarrow f^{\prime }(x)dx=dt$
• Step 3: $I=\int \frac{f^{\prime }(x)}{f(x)}dx=\int \frac{1}{t}dt=\log |t|+C=\log |f(x)|+C$.

Note: If the numerator in integrand is the exact differential of thedenominator, then its integral is the logarithm of the denominator.

Some Standard Results:

1. $\int \tan xdx=-\log |\cos x|+C$
2. $\int \cot xdx=\log |\sin x|+C$
3. $\int \sec xdx=\log |\sec x+\tan x|+C$
4. $\int \mathrm{cosec}xdx=\log |\mathrm{cosec}x-\cot x|+C$.

Type IV: Integrals of the form, $\int [f(x){]}^n{f}^{\prime }(x)dx$:

Formula: $\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1},n\ne -1$

• Step 1: Let $I=\int [f(x){]}^n{f}^{\prime }(x)dx$
• Step 2: Substitute $f(x)=t\Rightarrow f^{\prime }(x)dx=dt$
• Step 3: $I=\int (f(x){)}^n{f}^{\prime }(x)dx=\int t^{n}dt=\frac{t^{n+1}}{n+1}+C=\frac{[f(x){]}^{n+1}}{n+1}+C,n\ne -1$.

Type V: Evaluation of Integrals by Using the Trigonometric Substitutions:

Expression	Substitution
$a^2+x^2$	$x=a\tan \theta$ or $a\cot \theta$
$a^2-x^2$	$x=a\sin \theta$ or $a\cos \theta$
$x^2-a^2$	$x=a\sec \theta$ or $a\mathrm{cosec}\theta$
$\sqrt{\frac{a-x}{a+x}}$ or, $\sqrt{\frac{a+x}{a-x}}$	$x=a\cos 2\theta$
$\sqrt{\frac{x-\alpha }{\beta -x}}$ or $\sqrt{(x-\alpha )(x-\beta )}$	$x=\alpha {\cos }^2\theta +\beta {\sin }^2\theta$

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Following are some substitutions useful in evaluating integrals,

Some Special Integrals:

1. $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$
2. $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$
3. $\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$
4. $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+C$
5. $\int \frac{1}{\sqrt{a^2+x^2}}dx=\log |x+\sqrt{a^2+x^2}|+C$
6. $\int \frac{1}{\sqrt{x^2-a^2}}dx=\log |x+\sqrt{x^2-a^2}|+C$

Type VI : Integrals of the form, $\int \frac{px+q}{a{x}^2+bx+c}dx$

• Step 1: Write $px+q=\lambda \left\{\frac{d}{dx}(a{x}^2+bx+c)\right\}+\mu$, i.e. $px+q=\lambda (2ax+b)+\mu$
• Step 2: Obtain the values of $\lambda$ and $\mu$ by equating the coefficients of like powers of $x$ on both sides.
• Step 3: Replace $px+q$ by $\lambda (2ax+b)+\mu$ in the given interval to get
  $\int \frac{px+q}{a{x}^2+bx+c}dx=\lambda \int \frac{2ax+b}{a{x}^2+bx+c}dx+\mu \int \frac{1}{a{x}^2+bx+c}dx$
• Step 4: Integrate RHS in step $3$ and put the values of $\lambda$ and $\mu$ obtained in step $2$.
• Step 5: The first integral on RHS can be solved using substituting method, and the second integral can be solved using the following formulae.
  $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C,\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C$, and
  $\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$

Integration Using Trigonometric Identities

While integrating a function with a trigonometric integrand, we employ trigonometric identities and simplify the function.

Integrals of the Form, $\int \frac{1}{a\sin x+b\cos x}dx,\int \frac{1}{a+b\sin x}dx,\int \frac{1}{a+b\cos x}dx,\int \frac{1}{a\sin x+b\cos x+c}dx$

• Step 1: Put $\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$, and $\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$ and simplify.
• Step 2: Replace $1+{\tan }^2\frac{x}{2}$ in the numerator by ${\sec }^2\frac{x}{2}$.
• Step 3: Put $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$. This substitution reduces the integral in the form $\int \frac{1}{a{t}^2+bt+c}dt$.
• Step 4: Evaluate integral obtained in step $3$ by converting it into the standard form whose formula is known to us.

Few more trigonometric identities which are used in simplifying the expression of integrand are as follows:

• ${\sin }^{2}x=\frac{1-\cos 2x}{2}$
• ${\cos }^{2}x=\frac{1+\cos 2x}{2}$
• ${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4}$
• ${\cos }^{3}x=\frac{3\cos x+\cos 3x}{4}$

Integration by Parts

If $u$ and $v$ are two functions of $x$, then

$\int uvdx=u(\int vdx)-\int \{\frac{du}{dx}\int vdx\}dx$

The integral of the product of two functions $=$ (First function) $\times$ (Integral of second function) $-$ Integral of $\{$(Differentiation of first function) $\times$ (integral of second function)$\}$

In the above rule, we can also choose the first function as the function which comes first in the word ILATE, where

$I-$ Inverse trigonometric function
$L-$ Logarithmic function
$A-$ Algebraic function
$T-$ Trigonometric function
$E-$ Exponential function

Integrals of the form, $\int e^x[f(x)+f^{\prime }(x)]dx$

Formula: $\int e^x\{f(x)+f^{\prime }(x)\}dx=e^{x}f(x)+C$

• Step 1: Express the given integral as the sum of two integrals, one consisting of $f(x)$, and other containing $f^{\prime }(x)$.
  i. e. $\int e^x\{f(x)+f^{\prime }(x)\}dx=\int e^{x}f(x)dx+\int e^x{f}^{\prime }(x)dx$
• Step 2: Evaluate the first integral on RHS using integration by parts by taking $e^x$ as the second function. The second integral on RHS will cancel out from the second term obtained by evaluating the first integral.

Note: The above formula is also true if we have $e^{kx}$ in place of $e^x$ i.e., $\int e^{kx}\{kf(x)+f^{\prime }(x)\}dx=$ $e^{kx}f(x)+C$.

Some Important Integrals:

(i) $\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\left(\frac{x}{a}\right)+C$

(ii) $\int \sqrt{a^2+x^2}dx=\frac{1}{2}x\sqrt{a^2+x^2}+\frac{1}{2}{a}^2\log |x+\sqrt{a^2+x^2}|+C$

(iii) $\int \sqrt{x^2-a^2}dx=\frac{1}{2}x\sqrt{x^2-a^2}-\frac{1}{2}{a}^2\log |x+\sqrt{x^2-a^2}|+C$

Integration of Some Particular Functions

1. Integrals of the Form, $\int \sqrt{a{x}^2+bx+c}dx$

• Step 1: Make coefficient of $x^2$ as one by taking ‘$a$’ common to obtain $x^2+\frac{b}{a}x+\frac{c}{a}$.
• Step 2: Add and subtract ${\left(\frac{b}{2a}\right)}^2$ in $x^2+\frac{b}{a}x+\frac{c}{a}$ to obtain ${(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2}$
  After applying these two steps the integral reduces to one of the following three forms:
  $\int \sqrt{a^2+x^2}dx,\int \sqrt{a^2-x^2}dx,\int \sqrt{x^2-a^2}dx$
• Step 3: Use the appropriate formula.

2. Integrals of the Form, $\int (px+q)\sqrt{a{x}^2+bx+c}dx$

In order to evaluate the above type of integrals, we use the following algorithm,

• Step 1: Express $px+q$ as $px+q=\lambda \frac{d}{dx}(a{x}^2+bx+c)+\mu$ i.e., $px+q=\lambda (2ax+b)+\mu$.
• Step 2: Obtain the values of $\lambda$ and $\mu$ by equating the coefficients of $x$ and constant terms on both sides.
• Step 3: Replace $px+q$ by $\lambda (2ax+b)+\mu$ in the integral to obtain
  $\int (px+q)\sqrt{a{x}^2+bx+c}dx=\lambda \int (2ax+b)\sqrt{a{x}^2+bx+c}dx+\mu \int \sqrt{a{x}^2+bx+c}dx$
• Step 4: To evaluate first integral on RHS, use the formula $\int (f(x){)}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}$. And, the second integral is evaluated by using the method discussed above.

3. Integrals of the Form, $\int \frac{1}{\sqrt{a{x}^2+bx+c}}dx$

To evaluate this type of integrals, use the following algorithm.

• Step 1: Make the coefficient of $x^2$ unity, if it is not.
• Step 2: Find half of the coefficient of $x$.
• Step 3: Add and subtract ${\left(\frac{1}{2}\text{ Coefficient of }x\right)}^2$ inside the square root to express the quantity in the form ${(x+\frac{b}{2a})}^2+\frac{4ac-b^2}{4a^2}$ or, $\frac{4ac-b^2}{4a^2}-{(x+\frac{b}{2a})}^2$.
• Step 4: Use the suitable formula from the following formulas:
  $\int \frac{1}{\sqrt{a^2+x^2}}dx=\log |x+\sqrt{a^2+x^2}|+C,\int \frac{1}{\sqrt{x^2-a^2}}dx=\log |x+\sqrt{x^2-a^2}|+C$ and $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+C$.

Integration by Using Partial Fractions

Any proper rational function $\frac{f(x)}{g(x)}$ can be expressed as the sum of two rational functions, each having a simple factor of $g(x)$. Each such fraction is called a partial fraction. The process of obtaining them is called the resolution or decomposition of $\frac{f(x)}{g(x)}$ into partial fractions.

The resolution of $\frac{f(x)}{g(x)}$ into partial fractions depends mainly upon the nature of the factors of $g(x)$ as discussed below:

Case I :  When $g(x)$ contains non-repeated linear factors only

Let $g(x)=(x-a_1)(x-a_2)\dots (x-a_n)$. Then, we assume that

$\frac{f(x)}{g(x)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+\cdots +\frac{A_{\pi }}{x-a_n}$

where, $A_1,A_2,\dots ,A_n$ are constants.

Step to determine the constants:

• Step 1.Take the LCM on RHS.
• Step 2. Equate the numerator on RHS to the numerator on LHS
• Step 3. Substitute, $x=a_1,a_2,\dots ,a_n$

Case II : When $g(x)$ contains some repeated linear factors and the remaining are non-repeated linear factors

Let $g(x)=(x-a{)}^k(x-a_1)(x-a_2)\dots (x-a_r)$. Then, we assume that

$\frac{f(x)}{g(x)}=\frac{A_1}{x-a}+\frac{A_2}{(x-a{)}^2}+\frac{A_3}{(x-a{)}^3}+\cdots +\frac{A_k}{(x-a{)}^k}+\frac{B_1}{x-a_1}+\frac{B_2}{x-a_2}+\cdots +\frac{B_r}{(x-a_r)}$

i.e., corresponding to non-repeating factors we assume as in Case $I$ and for each repeating factor $(x-a{)}^k$, we assume partial fractions $\frac{A_1}{x-a}+\frac{A_2}{(x-a{)}^2}+\frac{A_3}{(x-a{)}^3}+\cdots +\frac{A_k}{(x-a{)}^k}$, where $A_1,A_2,\dots ,A_k$ are constants.

Now, to determine constants we equate numerators on both sides. Some of the constants are determined by comparing coefficients of equal powers of $x$ on both sides.

Case III: When $g(x)$ contains non-repeated irreducible factors of the form $a{x}^2+bx+c$

Corresponding to each quadratic factor of $a{x}^2+bx+c$, we assume partial fraction of the type $\frac{Ax+B}{a{x}^2+bx+c}$, where $A$ and $B$ are constants to be determined by comparing coefficients of similar powers of $x$ in the numerator of both sides.

Case IV: When $g(x)$ contains repeated and non-repeated irreducible quadratic factors of the form ${(a{x}^2+bx+c)}^n$

Corresponding to every repeated irreducible quadratic factor of $g(x)$ there exist partial fractions of the form $\frac{A_{1}x+B_1}{(\mathit{a}\mathit{x}{}^2+\mathit{b}\mathit{x}+\mathit{c})},\frac{A_{2}x+B_2}{{(\mathit{a}{x}^2+\mathit{b}\mathit{x}+\mathit{c})}^2},\cdots ,\frac{A_{n}x+B_n}{{(\mathit{a}\mathit{x}{}^2+\mathit{b}x+c)}^n}$, where $A_1,A_2,\dots \dots ,A_n$, and $B_1,B_2,\dots ,B_n$ are real numbers.

Solved Examples – Methods of Integration

Q.1. Evaluate: $\int \frac{x^3+x+1}{x^2-1}dx$
Ans: Let $I=\int \frac{x^3+x+1}{x^2-1}dx$
$\Rightarrow \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}$
$\therefore I=\int (x+\frac{2x+1}{x^2-1})dx$
$\Rightarrow I=\int xdx+\int \frac{2x}{x^2-1}dx+\int \frac{1}{x^2-1}dx$$\dots \dots .(i)$
Let $x^2-1=t$
$\Rightarrow 2xdx=dt$
Thus, $\int \frac{2x}{x^2-1}dx=\int \frac{1}{t}dt=\log |t|+C=\log |x^2-1|+C_1\dots \dots (ii)$
Also, we know that $\int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C_2$
$\therefore \int \frac{1}{x^2-1}dx=\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C_2\dots \dots .(iii)$
$\Rightarrow \int xdx=\frac{x^2}{2}+C_3\dots \dots .(iv)$
Substituting the values of integral from $(ii),(iii)$, and $(iv)$, in $(i)$ we have,
$I=\frac{x^2}{2}+\log |x^2-1|+\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C$

Q.2. Evaluate: $\int \frac{2x^3}{4+x^8}dx$
Ans: Let $I=\int \frac{2x^3}{4+x^8}dx=\int \frac{2x^3}{2^2+{\left(x^4\right)}^2}dx$
$x^4=t$
$\Rightarrow 4x^{3}dx=dt$
$\Rightarrow dx=\frac{dt}{4x^3}$
$\therefore I=\int \frac{2x^3}{4+t^2}\times \frac{dt}{4x^3}=\frac{1}{2}\int \frac{1}{2^2+t^2}dt$
$\Rightarrow I=\frac{1}{2}\times \frac{1}{2}{\tan }^{-1}\left(\frac{t}{2}\right)+C$
$\therefore \int \frac{2x^3}{4+x^8}dx=\frac{1}{4}{\tan }^{-1}\left(\frac{x^4}{2}\right)+C$

Q.3. Evaluate: $\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx$
Ans: Let $I=\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx$
Dividing the numerator and denominator of the given integrand by ${\cos }^{2}x$, we get
$\Rightarrow I=\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx=\int \frac{{\sec }^{2}x}{a^2{\tan }^{2}x+b^2}dx$
Let $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt$
Thus,
$I=\int \frac{1}{a^2{t}^2+b^2}dt$
$=\frac{1}{a^2}\int \frac{dt}{t^2+{\left(\frac{b}{a}\right)}^2}$
$=\frac{1}{a^2}\times \frac{1}{\frac{b}{a}}{\tan }^{-1}\left(\frac{t}{\frac{b}{a}}\right)+C$
$=\frac{1}{ab}{\tan }^{-1}\left(\frac{at}{b}\right)+C$
$\therefore \int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx=\frac{1}{ab}{\tan }^{-1}\left(\frac{a\tan x}{b}\right)+C$

Q.4. Evaluate: $\int \frac{1}{2-3\cos 2x}dx$
Ans: Let $I=\int \frac{1}{2-3\cos 2x}dx$
$\Rightarrow I=\int \frac{1}{2-3({\cos }^{2}x-{\sin }^{2}x)}dx$
$I=\int \frac{{\sec }^{2}x}{2{\sec }^{2}x-3+3{\tan }^{2}x}dx$ [Dividing the numerator and denominator by ${\cos }^{2}x$ ]
$=\int \frac{{\sec }^{2}x}{2(1+{\tan }^{2}x)-3+3{\tan }^{2}x}dx$
$I=\int \frac{{\sec }^{2}x}{5{\tan }^{2}x-1}dx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$\therefore I=\int \frac{1}{5t^2-1}dt$
$=\frac{1}{5}\int \frac{dt}{t^2-{\left(\frac{1}{\sqrt{5}}\right)}^2}$
$=\frac{1}{5}\times \frac{1}{2\left(\frac{1}{\sqrt{5}}\right)}\log \left|\frac{t-\frac{1}{\sqrt{5}}}{t+\frac{1}{\sqrt{5}}}\right|+C$
$\therefore \int \frac{1}{2-3\cos 2x}dx=\frac{1}{2\sqrt{5}}\log \left|\frac{\sqrt{5}\tan x-1}{\sqrt{5}\tan x+1}\right|+C$

Q.5. Evaluate: $\int e^{ax}\sin (bx+c)dx$
Ans: Let $I=\int e^{ax}\sin (bx+c)dx$
Taking first function as $\sin (bx+c)$, and second function as $e^{ax}$ then using the formula of Integrating by parts, i.e., $\int uvdx=u(\int vdx)-\int \{\frac{du}{dx}\int vdx\}dx$, we get
Thus,
$I=\sin (bx+c)\times \int e^{ax}dx-\int [\frac{d}{dx}(\sin (bx+c))\times \int e^{ax}dx]dx$
$=\sin (bx+c)\frac{e^{ax}}{a}-\int b\cos (bx+c)\cdot \frac{e^{ax}}{a}dx$
$=\sin (bx+c)\frac{e^{ax}}{a}-\frac{b}{a}\int \cos (bx+c)\cdot e^{ax}dx\dots \dots (i)$
Consider,
$I_1=\int \cos (bx+c)\cdot e^{ax}\cdot dx$
Again applying integration by parts, we get
$I_1=\cos (bx+c)\times \frac{e^{ax}}{a}-\int [-b\sin (bx+c)\times \frac{e^{ax}}{a}]dx$
$=\cos (bx+c)\times \frac{e^{ax}}{a}+\frac{b}{a}\int \sin (bx+c)\times e^{ax}dx$
$=\cos (bx+c)\times \frac{e^{ax}}{a}+\frac{b}{a}I$
Substituting $I_1$ in $(i)$, we have
$I=\sin (bx+c)\frac{e^{ax}}{a}-\frac{b}{a}[\cos (bx+c)\times \frac{e^{ax}}{a}+\frac{b}{a}I]$
$=\sin (bx+c)\frac{e^{ax}}{a}-\frac{b}{a^2}\times e^{ax}\times \cos (bx+c)-\frac{b^2}{a^2}I$
$\Rightarrow I+\frac{b^2}{a^2}I=\sin (bx+c)\frac{e^{ax}}{a}-\frac{b}{a^2}\times e^{ax}\cos (bx+c)$
Hence, $I=\frac{e^{ax}}{(1+\frac{b^2}{a^2})}(\frac{\sin (bx+c)}{a}-\frac{b}{a^2}\cos (bx+c))$

Summary of Methods of Integration

Many integration formulas can be simply deduced from their derivative formulas, however some integration problems take more effort. In order to integrate various functions, we follow the methods like substitution and change of variables, integration by parts, trigonometric integrals, and integration using partial fractions. In this article, we have discussed all the methods of integration in brief so that they can be easily understood. We have provided some standard formulae like $\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C,\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{\{f(x){\}}^{n+1}}{n+1},n\ne -1,\int e^x\{f(x)+f^{\prime }(x)\}dx=e^{x}f(x)+C$ etc. There are also provided few solved examples, and FAQs which are the applications of all discussed methods.

Frequently Asked Questions (FAQs) on Methods of Integration

Q.1. How many methods of integration are there?
Ans: There are five methods of integration:
1. Integration by Substitution
2. Integration Using Trigonometric Identities.
3. Integration by Parts
4. Integration of Some Particular Functions
5. Integration Using Partial Fractions

Q.2. What are the $5$ basic integration formulas?
Ans: The five basic integration formulas are as follows:
1. $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1$
2. $\int \frac{1}{x}dx={\log }_e|x|+C$
3. $\int e^{x}dx=e^x+C$
4. $\int a^{x}dx=\frac{a^x}{{\log }_{e}a}+C$
5. $\int e^x[f(x)+f^{\prime }(x)]dx$

Q.3. What is the ILATE Rule in Integration Techniques?
Ans: If $u$ and $v$ are two functions of $x$, then
$\int uvdx=u(\int vdx)-\int \{\frac{du}{dx}\int vdx\}dx$
In the above rule, we can also choose the first function as the function which comes first in the word ILATE, where:
$I-$ stands for the inverse trigonometric function.
$L-$ stands for the logarithmic function.
$A-$ stands for the algebraic function.
$T-$ stands for the trigonometric functions.
$E-$ stands for the exponential function.

Q.4. What is the proper partial function?
Ans: If $f(x)$ and $g(x)$ are polynomials, then $\frac{f(x)}{g(x)}$ defines a rational algebraic function or a rational function of $x$. If the degree of $f(x)<$ degree of $g(x)$, then $\frac{f(x)}{g(x)}$ is called a proper rational fraction.

Q.5. What is an improper partial function?
Ans: If $f(x)$ and $g(x)$ are polynomials, then $\frac{f(x)}{g(x)}$ defines a rational algebraic function or a rational function of $x$. If the degree of $f(x)\ge$ degree of $g(x)$, then $\frac{f(x)}{g(x)}$ is called an improper rational function.

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