The different methods of integration include:

• Integration by Substitution
• Integration by Parts
• Integration Using Trigonometric Identities
• Integration of Some particular function
• Integration by Partial Fraction

Integration by Substitution

Sometimes, it is really difficult to find the integration of a function, thus we can find the integration by substituting a new independent variable and this method is called Integration By Substitution.
The given form of integral function (say \(f(x)\) ) can be transformed into another by changing the independent variable \(x\) to \(t\),
By substituting \(x=g(t)\) in the function \(\int f(x)\), we get;
\({dx} / {dt}=g^{\prime}(\mathrm{t})\)
or \({dx}={g}^{\prime}({t}) \cdot {dt}\)
Thus, \(I=\int f(x) \cdot d x=f(g(t)) \cdot g^{\prime}(t) \cdot d t\)

Example 1: Evaluate \(\int \cos \left(x^2\right) 2 x d x\)

Solution:
\(\int \cos \left(x^2\right) 2 x d x\)
We know (from above) that it is in the right form to do the substitution:
let’s substitute \(u={x}^2\)
\(
\begin{aligned}
&\int \cos \left(x^2\right) 2 x d x \\
&=\int \cos (u) d u
\end{aligned}
\)
Now integrate:
\(
\int \cos (u) d u=\sin (u)+C
\)
And finally put \({u}={x}^2\) back again:
\(
=\sin \left(x^2\right)+C
\)
\(
\text { So } \int \cos \left(x^2\right) 2 x d x=\sin \left(x^2\right)+C
\)

Integration by Parts

Integration by Parts is a special method of integration that is often useful when two functions are multiplied together but is also helpful in other ways. Let us see the rule:

\(
\int u v d x=u \int v d x-\int u^{\prime}\left(\int v d x\right) d x
\)
\({u}\) is the function \(u(x)\)
\({v}\) is the function \({v}({x})\)
\({u}^{\prime}\) is the derivative of the function \(u(x)\)

Example 2: Evaluate \(\int x \cos (x) d x?\)

Solution:
we have \({x}\) multiplied by \(\cos ({x})\), so integration by parts is a good choice.
First choose which functions for \({u}\) and \({v}\) :
\({u}={x}\)
\(v=\cos (x)\)
So now it is in the format \(\int {u} {v} {d} {x}\) we can proceed:
Differentiate \({u}: {u}^{\prime}={x}^{\prime}=1\)
Integrate \({v}: \int v {dx}=\int \cos ({x}) {dx}=\sin ({x}) \quad\) (see Integration Rules )
Now we can put it together:
Simplify and solve:
\(
\begin{aligned}
\int x \cos (x) d x =x \sin (x)-\int \sin (x) d x =x \sin (x)+\cos (x)+C
\end{aligned}
\)

Example 3: What is \(\int \ln (x) / x^2 d x ?\)

Solution:
First choose \(u\) and \(v\) :
\(\mathrm{u}=\ln (\mathrm{x})\)
\(v=1 / x^2\)
Differentiate \(u: \ln (x)^{\prime}=\frac{1}{x}\)
Integrate \(\mathrm{v}: \int 1 / \mathrm{x}^2 \mathrm{dx}=\int \mathrm{x}^{-2} \mathrm{dx}=-\mathrm{x}^{-1}=\frac{-1}{\mathrm{x}}\) (by the power rule)
\(
\begin{aligned}
\int \ln (x) / x^2 d x =-\ln (x) / x-\int-1 / x^2 d x \\
=-\ln (x) / x-1 / x+C \\
=-\frac{\ln (x)+1}{x}+C
\end{aligned}
\)

Integration using Trigonometric Identities

In the integration of a function, if the integrand involves any kind of trigonometric function, then we use trigonometric identities to simplify the function that can be easily integrated. A few of the trigonometric identities are as follows:

\(\sin (2 x)=2 \sin (x) \cos (x)\)
\(\cos (2 x)=\cos ^2(x)-\sin ^2(x)=2 \cos ^2(x)-1=1-2 \sin ^2(x)\)
\(
\int \tan ^2(x) d x=\int\left(\sec ^2(x)-1\right) d x=\tan (x)-x+c
\)
\(
\begin{aligned}
&\sin (A)+\sin (B)=2 \sin \frac{A+B}{2} \cos \frac{A-B}{2} \\
&\sin (A)-\sin (B)=2 \cos \frac{A+B}{2} \sin \frac{A-B}{2} \\
&\cos (A)+\cos (B)=2 \cos \frac{A+B}{2} \cos \frac{A-B}{2} \\
&\cos (A)-\cos (B)=-2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}
\end{aligned}
\)
\(
\begin{aligned}
&\sin ^2 x=\frac{1-\cos 2 x}{2} \\
&\cos ^2 x=\frac{1+\cos 2 x}{2} \\
&\sin ^3 x=\frac{3 \sin x-\sin 3 x}{4} \\
&\cos ^3 x=\frac{3 \cos x+\cos 3 x}{4}
\end{aligned}
\)
All these identities simplify integrand, which can be easily found out.

Example 4: Find \(\int \sin 2 x \cos 3 x d x\)

Solution:

Given: \(\int \sin 2 x \cos 3 x d x\)
Now, by using the trigonometric identity \(\sin x \cos y=\left(\frac{1}{2}\right)[\sin (x+y)+\sin (x-y)]\)
Therefore, \(\int \sin 2 x \cos 3 x d x=\left(\frac{1}{2}\right)\left[\int \sin 5 x d x-\int \sin x d x\right]\)
\(=(1 / 2)[(-1 / 5 \cos 5 x+\cos x]+C\)
\(=(-1 / 10) \cos 5 x+(1 / 2) \cos x+C\)
Therefore, \(\int \sin 2 x \cos 3 x d x=(-1 / 10) \cos 5 x+(1 / 2) \cos x+C\).

Example 5: \(\int \sin ^3 x \cos ^2 x d x\)

Solution:

Given: \(\int \sin ^3 x \cos ^2 x d x\)
\(\int \sin ^3 x \cos ^2 x d x=\int \sin ^2 x \sin x \cos ^2 x d x\)
We know that \(\sin ^2 x=1-\cos ^2 x\).
Therefore,
\(\int \sin ^3 x \cos ^2 x d x=\int\left(1-\cos ^2 x\right) \cos ^2 x \sin x d x\)
Now, put \(t=\cos x\), and hence \(d t=-\sin x d x\)
Hence,
\(
\begin{aligned}
&\int \sin ^2 x \cos ^2 x \sin x d x=-\int\left(1-t^2\right) t^2 d t \\
&=-\int\left(t^2-t^4\right) d t
\end{aligned}
\)
Now, integrating the above function, we get
\(\int \sin ^2 x \cos ^2 x \sin x d x=-\left[\left(t^3 / 3\right)-\left(t^5 / 5\right)\right]+C\)
Now, again substitute \(t=\cos x\), we get
\(\int \sin ^2 x \cos ^2 x \sin x d x=-(1 / 3) \cos ^3 x+(1 / s) \cos ^5 x+C\)
Therefore, \(\int \sin ^3 x \cos ^2 x d x=-(1 / 3) \cos ^3 x+(1 / 5) \cos ^5 x+C\).

Example 6: Evaluate: \(\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x\)

Solution:

\(\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x\)
Dividing the numerator and denominator of the given integrand by \(\cos ^2 x\), we get \(I=\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\int \frac{\sec ^2 x}{a^2 \tan ^2 x+b^2} d x\)
Putting \(\tan x=t\) and \(\sec ^2 x d x=d t\), we get
\(
\begin{aligned}
&I=\int \frac{d t}{a^2 t^2+b^2} \\
&=\frac{1}{a^2} \int \frac{d t}{t^2+\left(\frac{b}{a}\right)^2} \\
&=\frac{1}{a^2} \times \frac{1}{\frac{b}{a}} \tan ^{-1}\left(\frac{t}{\frac{b}{a}}\right)+C \\
&=\frac{1}{a b} \tan ^{-1}\left(\frac{a t}{b}\right)+C \\
&=\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+C \\
&\text { Hence, } \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+C
\end{aligned}
\)

Example 7: Evaluate: \(\int \sin ^3 x \cos ^4 x d x\)

Solution:
\(I=\int \sin ^3 x \cos ^4 x d x\)
Here,power of \(\sin x\) is odd, so we substitute \(\cos x=t\) \(\Rightarrow-\sin x d x=d t \Rightarrow d x=-\frac{d t}{\sin x}\)
\(\therefore I=\int \sin ^3 x t^4\left(-\frac{d t}{\sin x}\right)\)
\(=-\int\left(\sin ^2 x \times t^4\right) d t\)
\(=-\int\left(1-t^2\right) t^4 d t\)
\(=-\int\left(t^4-t^6\right) d t\)
\(\Rightarrow I=-\frac{t^5}{5}+\frac{t^7}{7}+C\)
Hence, \(\int \sin ^3 x \cos ^4 x d x=-\frac{\cos ^5 x}{5}+\frac{\cos ^7 x}{7}+C\)

Example 8: Evaluate: \(\int \frac{\sin ^4 x}{\cos ^8 x} d x\)

Solution:
\(I=\int \frac{\sin ^4 x}{\cos ^8 x} d x\)
\(=\int\left(\frac{\frac{\sin ^4 x}{\cos ^4 x}}{\frac{\cos ^8 x}{\cos ^4 x}}\right) d x\)
\(=\int\left(\tan ^4 x \sec ^4 x\right) d x\)
\(=\int\left\{\tan ^4 x\left(1+\tan ^2 x\right) \sec ^2 x\right\} d x\)
Putting \(\tan x=t\), and \(\sec ^2 x d x=d t\) we get
\(
\begin{aligned}
&I=\int t^4\left(1+t^2\right) d t \\
&=\frac{t^5}{5}+\frac{t^7}{7}+C
\end{aligned}
\)
Substituting for \(t\), we get
\(
\therefore I=\frac{\tan ^5 x}{5}+\frac{\tan ^7 x}{7}+C
\)
Hence, \(\int \frac{\sin ^4 x}{\cos ^8 x} d x=\frac{\tan ^5 x}{5}+\frac{\tan ^7 x}{7}+C\)

Example 9: Evaluate: \(\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x\)

Solution:
\(I=\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x\)
\(
\begin{aligned}
&=\int \frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^4 x-\cos ^4 x\right)}{\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x} d x \\
&=\int \frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^2 x-\cos ^2 x\right)}{\sin ^4 x+\cos ^4 x} d x \\
&=-\int \cos 2 x d x\left[\because \cos ^2 x-\sin ^2 x=\cos 2 x\right] \\
&=-\frac{1}{2} \sin 2 x+C
\end{aligned}
\)
Hence, \(\int \frac{\sin ^3 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=-\frac{1}{2} \sin 2 x+C\)

Example 10: Evaluate: \(\int \sin ^4 x d x\)

Solution:

Let \(I=\int \sin ^4 x d x\)
\(
\begin{aligned}
&=\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x\left[\because \sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right] \\
&=\frac{1}{4} \int\left(1-2 \cos 2 x+\cos ^2 2 x\right) d x \\
&=\frac{1}{4} \int\left(1-2 \cos 2 x+\frac{1+\cos 4 x}{2}\right) d x \\
&=\frac{1}{8} \int(2-4 \cos 2 x+1+\cos 4 x) d x \\
&=\frac{1}{8} \int(3-4 \cos 2 x+\cos 4 x) d x \\
&=\frac{1}{8}\left\{3 x-2 \sin 2 x+\frac{\sin 4 x}{4}\right\}+C
\end{aligned}
\)
Hence, \(\int \sin ^4 x d x=\frac{1}{8}\left\{3 x-2 \sin 2 x+\frac{\sin 4 x}{4}\right\}+C\)

Integration of some particular function

Integration of some particular function involves some important formulae of integration that can be applied to make other integration into the standard form of the integrand. The integration of these standard integrands can be easily found using a direct form of integration method.

Example 11: Evaluate \(
\int \frac{{dx}}{{x}^2-{a}^2}=\frac{1}{2 {a}} \log \left|\frac{{x}-{a}}{{x}+{a}}\right|+\mathbf{C}\)

Solution:
The integral function can be split into the sums of partial fractions, i.e.
\(
\int \frac{d x}{x^2-a^2}=\int \frac{d x}{(x-a)(x+a)}=\int \frac{A}{(x-a)} \cdot d x+\int \frac{B}{(x+a)} \cdot d x \dots(i)
\)
Solving for values of \(A\) and \(B\), we have,
\(1=A(x+a)+B(x-a)\)
Putting \(x=a\) and then \(-a\), we get the values of \(A\) and \(B\) to be
\(
\begin{aligned}
&\frac{1}{2 a} \text { and } -\frac{1}{2 a} \text { respectively. }
\end{aligned}
\)
Substituting these values in (i), we have
\(
\begin{aligned}
&\int \frac{d x}{x^2-a^2}=\int \frac{d x}{2 a(x-a)}-\int \frac{d x}{2 a(x+a)} \\
&=\frac{1}{2 a}\left[\int \frac{d x}{(x-a)}-\int \frac{d x}{(x+a)}\right] \\
&=\frac{1}{2 a}[\log |x-a|-\log |x+a|]+C \\
&=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C
\end{aligned}
\)

Integration by Partial Fraction

We know that a Rational Number can be expressed in the form of \(p / q\), where \(p\) and \(q\) are integers and \(q \neq 0\). Similarly, a rational function is defined as the ratio of two polynomials which can be expressed in the form of partial fractions: \(P(x) / Q(x)\), where \(Q(x) \neq 0\).

There are in general two forms of partial fraction:

• Proper partial fraction: When the degree of the numerator is less than the degree of the denominator, then the partial fraction is known as a proper partial fraction.
• Improper partial fraction: When the degree of the numerator is greater than the degree of the denominator then the partial fraction is known as an improper partial fraction. Thus, the fraction can be simplified into simpler partial fractions, that can be easily integrated.

Suppose we have to find \(y=\int \frac{P(x)}{Q(x)} d x\) where \(\frac{P(x)}{Q(x)}\) is an improper rational function. We reduce it in such a way that \(\frac{P(x)}{Q(x)}=T(x)+\frac{P_1(x)}{Q(x)}\) . Here, \(T(x)\) is polynomial in \(x\) and \(\frac{P_1(x)}{Q(x)}\) is proper rational function. The following table shows some rational functions and their corresponding form of partial fractions.

\(
\begin{array}{|c|c|}
\hline \text { Rational Function } & \text { Form of Partial Function } \\
\hline \frac{p x+q}{(x-a)(x-b)}, \text { where } a \neq b & \frac{A}{(x-a)}+\frac{B}{(x-b)} \\
\frac{p x+q}{(x-a)^2} & \frac{A}{(x-a)}+\frac{B}{(x-a)^2} \\
\frac{p x^2+q x+r}{(x-a)(x-b)(x-c)} & \frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{A}{(x-c)} \\
\frac{p x^2+q x+r}{(x-a)^2(x-b)} & \frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{(x-b)} \\
\frac{p x^2+q x+r}{(x-a)\left(x^2+b x+c\right)} & \frac{A}{(x-a)}+\frac{B x+c}{x^2+b x+c} \\
\hline
\end{array}
\)

 

Example 12: Find the integral of \(f(x)=\frac{1}{(x+1)(x+2)}\) using integration by partial fractions.

Solution:

By using partial fraction we have \(\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2} \cdots(1)\).
We will determine the values of \(A\) and \(B\).
On comparing in equation (1), we get \(1=\mathrm{A}(\mathrm{x}+2)+\mathrm{B}(\mathrm{x}+1)\).
From this, we have a set of two linear equations.
\(A+B=0\) and \(2 A+B=1\)
On solving these equations we get, \(A=1\) and \(B=-1\).
So, equation (1) can be written as \(\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}\).
Now, solving the integral
\(
\begin{array}{r}
\quad \int\left(\frac{1}{(x+1)(x+2)}\right) d x \\
=\int\left(\frac{1}{x+1}-\frac{1}{x+2}\right) d x \\
=\log |x+1|-\log |x+2|+C \\
=\log \left|\frac{x+1}{x+2}\right|+C
\end{array}
\)

Example 13: \(\text { Find the integral } \int \frac{x^2+1}{x^2-5 x+6} d x\)

Solution:

Observe that the function \(\frac{x^2+1}{x^2-5 x+6}\) is an improper rational function.
Thus we apply integration using partial fractions.
On dividing this function, we get \(5 x-5\) as the remainder.
So, by long division, this can be written as \(\frac{x^2+1}{x^2-5 x+6}=1+\frac{5 x-5}{x^2-5 x+6}\)
By factorization, we have \(x^2-5 x+6=(x-2)(x-3)\).
So, \(1+\frac{5 x-5}{x^2-5 x+6}=1+\frac{5 x-5}{(x-2)(x-3)}\)
Let’s proceed with partial fraction on \(\frac{5 x-5}{(x-2)(x-3)}\).
\(\frac{5 x-5}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}\)
On comparing, we get, \(5 x-5=A(x-3)+B(x-2)\).
From this, we have a set of two linear equations.
\(
A+B=5 \cdots \cdots(1)
\)
\(
3 A+2 B=5 \cdots(2)
\)
On solving these equations (1) and (2) we get, \(A=-5\) and \(B=10\)
\(
1+\frac{5 x-5}{x^2-5 x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}
\)
Integrate both sides of the equation,
\(
\begin{aligned}
I &=\int\left(1+\frac{5 x-5}{x^2-5 x+6}\right) d x \\
&=\int\left(1-\frac{5}{x-2}+\frac{10}{x-3}\right) d x \\
&=\int d x-5 \int \frac{d x}{x-2}+10 \int \frac{d x}{x-3} \\
&=x-5 \log |x-2|+10 \log |x-3|+C
\end{aligned}
\)

Integrals of the Form

Integrals of the Form: \(\int \frac{f^{\prime}(x)}{f(x)} d x=\ln |f(x)|+C\)

Proof:

Step 1: Let \(I=\int \frac{f^{\prime}(x)}{f(x)} d x\).

Step 2: Substitute \(f(x)=t \Rightarrow f^{\prime}(x) d x=d t\)

Step 3: \(I=\int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{1}{t} d t=\ln |t|+C=\ln |f(x)|+C\).

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

Example 14: Evaluate \(\int \frac{3 x^2}{x^3+5} d x\)

Solution:

In this example we have \(f(x) = {x^3+5}\), derivative of \(f(x) = {3 x^2} = f^{\prime}(x)\). We can apply the formula \(\int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)+c\)  to solve this integral.

\(\int \frac{3 x^2}{x^3+1} d x=\ln \left(x^3+1\right)+c\)

 

Example 15: Evaluate \(\int \frac{e^x}{e^x+1} d x\)

Solution:

As the denominator derivative is equal to the numerator, we have \(\int \frac{e^x}{e^x+1} d x=\ln \left(e^x+1\right)+c\)

Example 16: Evaluate \(\int \tan x d x\)

Solution:

\(
\begin{aligned}
&\int \tan x d x \\
&=\int \frac{\sin x}{\cos x} d x \\
&=-\int \frac{-\sin x}{\cos x} d x \\
&=-\ln (\cos x)+c
\end{aligned}
\)

 

Example 17: Evaluate \(\int \frac{e^x+3 e^{-3 x}}{e^x-e^{-3 x}} d x\)

Solution: 

As it is in the form \(\int \frac{f^{\prime}(x)}{f(x)} d x\), answer is \(\ln({e^x-e^{-3 x}})+c\)