Reduction Formulas For Integration by Parts With Solved Examples

Reduction Formula

A reduction formula is the integration of the power of a function in terms of the integration of the lower power of the same function. By iteratively applying this formula, one can reduce a complicated integral into simpler forms, eventually leading to a basic integral that can be easily evaluated.

Derivation of a Reduction Formula

Proof for the integration of $\int x^n e^x$.

1. Integration by Parts:

Using the integration by parts formula, $d u \int u d v=u v-\int v$,

where: $u=x^n \quad$ and $\quad d v=e^x d x$

Therefore, $d u=n x^{n-1} d x$ and $v=e^x$

: $\int x^n e^x d x=x^n e^x-\int n x^{n-1} e^x d x$

(a) $\int \sin ^{\mathrm{n}} \mathrm{xdx}$

Let $\quad I_n=\int \sin ^n x d x=\int \sin ^{n-1} x \sin x d x$
take $\sin ^{\mathrm{n}-1} \mathrm{x}$ as first function and $\sin \mathrm{x}$ as second function

$
\begin{aligned}
& =-\sin ^{n-1} x \cos x+\int(n-1) \sin ^{n-2} x \cos ^2 x d x \\
& =-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x\left(1-\sin ^2 x\right) d x \\
& =-\sin ^{n-1} x \cos x+(n-1) \int\left(\sin ^{n-2} x-\sin ^n x\right) d x \\
& =-\sin ^{n-1} x \cos x+(n-1) I_{n-2}-(n-1) I_n \\
\therefore \quad n I_n & =-\sin ^{n-1} x \cos x+(n-1) I_{n-2} \\
\Rightarrow \quad I_n & =-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}
\end{aligned}
$
Thus, $\int \sin ^n x d x=\frac{-\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$

(b) $\int \tan ^n x d x$
$
\begin{aligned}
\text { Let } \quad I_n & =\int \tan ^n x d x \\
\Rightarrow \quad I_n & =\int \tan ^{n-2} x \tan ^2 x d x=\int \tan ^{n-2} x\left(\sec ^2 x-1\right) d x \\
& =\int \tan ^{n-2} x \sec ^2 x-I_{n-2}=\int t^{n-2} d t-I_{n-2}
\end{aligned}
$

where, $\tan x=t \Rightarrow \sec ^2 x d x=d t$

$\begin{aligned} & I_n=\frac{t^{n-1}}{n-1}-I_{n-2} \\ \therefore & I_n=\frac{\tan ^{n-1} x}{n-1}-I_{n-2} \\ \Rightarrow \quad & \int \tan ^n x d x=\frac{\tan ^{n-1} x}{n-1}-\int \tan ^{n-2} x d x\end{aligned}$

Integration of trigonometric function of power m for $m=4$.

$\therefore \int \tan ^4 x d x=\int \tan ^2 x \cdot \tan ^2 x d x=\int\left(\sec ^2 x-1\right) \tan ^2 x d x$Use $\sin ^4 x=\left(\sin ^2 x\right)^2=\left(\frac{1-\cos 2 x}{2}\right)^2, \cos ^4 x=\left(\cos ^2 x\right)^2=\left(\frac{1+\cos 2 x}{2}\right)^2$

(c) $\int \cos ^{\mathrm{n}} \mathrm{x} \mathrm{dx}$

Let $I_n=\int \cos ^n x d x=\int \cos ^{n-1} x \cos x d x$
Take $\cos ^{n-1} x$ as first function and $\cos x$ as second function.

$
\begin{aligned}
& =\cos ^{n-1} x \sin x+\int(n-1) \cos ^{n-2} x \sin ^2 x d x \\
& =\cos ^{n-1} x \sin x+(n-1) \int \cos ^{n-2} x\left(1-\cos ^2 x\right) d x \\
& =\cos ^{n-1} x \sin x+(n-1) I_{n-2}-(n-1) I_n \\
\therefore n I_n & =\cos ^{n-1} x \sin x+(n-1) I_{n-2} \\
\text { or } \int & \cos ^n x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x
\end{aligned}
$

(d) $\int \cot ^n \mathrm{x} \mathrm{dx}$

Let $I_n=\int \cot ^n x d x=\int \cot ^{n-2} x \cot ^2 x d x$

$
\begin{aligned}
& =\int \cot ^{n-2} x\left(\csc ^2 x-1\right) d x \\
& =\int \cot ^{n-2} x \csc ^2 x d x-I_{n-2} \\
& =\int t^{n-2} d t-I_{n-2}
\end{aligned}
$

where, $\cot x=t \Rightarrow \csc ^2 x d x=-d t$

$
\begin{aligned}
I_n & =-\frac{\cot ^{n-1} x}{n-1}-I_{n-2} \\
\therefore \quad \int \cot ^n x d x & =-\frac{\cot ^{n-1} x}{n-1}-\int \cot ^{n-2} x d x
\end{aligned}
$

(e) $\int \sec ^{\mathrm{n}} \mathrm{xdx}$

Let $I_n=\int \sec ^n x d x=\int \sec ^{n-2} x \sec ^2 x d x$
Take $\sec ^{\mathrm{n}-2} \mathrm{x}$ as first function and $\sec ^2 \mathrm{x}$ as a second function

$
\begin{aligned}
& \quad=\sec ^{n-2} x \tan x-\int(n-2) \sec ^{n-3} x \sec x \tan x \cdot \tan x d x \\
& \quad=\sec ^{n-2} x \tan x-(n-2) \int \sec ^{n-2} x\left(\sec ^2 x-1\right) d x \\
& \quad=\sec ^{n-2} x \tan x-(n-2) I_n+(n-2) I_{n-2} \\
& \Rightarrow(n-1) I_n=\sec ^{n-2} x \tan x+(n-2) I_{n-2} \\
& \text { or } \quad I_n=\frac{\sec ^{n-2} x \tan x}{(n-1)}+\frac{(n-2)}{(n-1)} I_{n-2} \\
& \therefore \int \sec ^n x d x=\frac{\sec ^{n-2} x \tan x}{(n-1)}+\frac{(n-2)}{(n-1)} \int \sec ^{n-2} x d x
\end{aligned}
$

(f) $\int \csc ^{\mathrm{n}} \mathrm{x} d \mathrm{x}$
$
\begin{aligned}
& \text { Let } I_n=\int \csc ^n x d x=\int \csc ^{n-2} x \csc ^2 x d x \\
& =\csc ^{n-2} x(-\cot x)-\int(n-2) \csc ^{n-2} x\left(\csc ^2 x-1\right) d x \\
& =-\csc ^{n-2} x(-\cot x)-(n-2) \int\left(\csc ^n x-\csc ^{n-2} x\right) d x \\
& =-\csc ^{n-2} x \cot x-(n-2) I_n+(n-2) I_{n-2} \\
& \therefore \quad(n-1) I_n=-\csc ^{n-2} x \cot x+(n-2) I_{n-2} \\
& \text { or } \quad I_n=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} I_{n-2} \\
& \therefore \int \csc ^n x d x=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \int \csc ^{n-2} x d x
\end{aligned}
$

Integration of the type $\int \cos ^{\mathrm{m}} \mathrm{x} \sin \mathrm{nx} \mathrm{dx}$

Let $\mathrm{I}_{\mathrm{m}, \mathrm{n}}=\int \cos ^{\mathrm{m}} \mathrm{x} \sin \mathrm{nx} \mathrm{dx}$
To evaluate this integral, we will use integration by parts method Here, take $\cos ^{\mathrm{m}} \mathrm{x}$ as the first function and $\sin \mathrm{nx}$ as the second function.

$
\begin{aligned}
& =-\frac{\cos ^m x \cos n x}{n}-\frac{m}{n} \int \cos ^{m-1} x \sin x \cos n x d x \\
& =-\frac{\cos ^m x \cos n x}{n}-\frac{m}{n} \int \cos ^{m-1} x\{\sin n x \cos x-\sin (n-1) x\} d x \\
& \text { [using } \sin (n-1) x=\sin n x \cos x-\cos n x \sin x \\
& \Rightarrow \sin x \cos n x=\sin n x \cos x-\sin (n-1) x] \\
& =-\frac{\cos ^m x \cos n x}{n}-\frac{m}{n} \int \cos ^m x \sin n x d x+\frac{m}{n} \int \cos ^{m-1} x \sin (n-1) x d x \\
& I_{m, n}=-\frac{\cos ^m x \cos n x}{n}-\frac{m}{n} I_{m, n}+\frac{m}{n} I_{m-1, n-1}
\end{aligned}
$

$\begin{aligned} & \Rightarrow \frac{m+n}{n} I_{m, n}=-\frac{\cos ^m x \cos n x}{n}+\frac{m}{n} I_{m-1, n-1} \\ & \text { or } \quad I_{m, n}=-\frac{\cos ^m x \cos n x}{m+n}+\frac{m}{m+n} I_{m-1, n-1}\end{aligned}$

NOTE :

In the similar way we can also prove the following result

1. $\int \cos ^m x \cos n x d x=\frac{\cos ^m x \sin n x}{m+n}+\frac{m}{m+n} \int \cos ^{m-1} x \cos (n-1) x d x$
2. $\int \sin ^m x \sin n x d x=\frac{n \sin ^m x \cos n x}{m^2-n^2}-\frac{m \sin ^{m-1} x \cos x \cos n x}{m^2-n^2}+\frac{m(m-1)}{m^2-n^2} \int \sin ^{m-2} x \sin n x d x$
3. $\int \sin ^m x \cos n x d x=\frac{n \sin ^m x \sin n x}{m^2-n^2}-\frac{m \sin ^{m-1} x \cos x \cos n x}{m^2-n^2}+\frac{m(m-1)}{m^2-n^2} \int \sin ^{m-2} x \cos n x d x$

Recommended Video Based on Integration by Reduction Formula

Solved Examples Based on Integration by Reduction Formula

Example 1: Evaluate $\int \sin ^4 x d x$

1) $\frac{1}{8}+\frac{\sin 4 x}{32}+\frac{\sin 2 x}{2}+C$
2) $\frac{\sin (4 x)-8 \sin (2 x)+12 x}{32}+C$
3) $\frac{1}{8}-\frac{\sin 4 x}{32}+\frac{\sin 2 x}{2}+C$

4) none of these

Solution
$
\begin{aligned}
& I=\int \sin ^4 x d x \\
& \Rightarrow I=\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x \\
& =\frac{1}{4} \int\left(1-2 \cos 2 x+\cos ^2 2 x\right) d x \\
& =\frac{1}{4} x-\frac{1}{2} \int \cos 2 x d x+\frac{1}{4} \int \cos ^2 2 x d x \\
& =\frac{1}{4} x-\frac{1}{4} \sin 2 x+\frac{1}{8}\left(x+\frac{1}{4} \sin (4 x)\right) \\
& =\frac{\sin (4 x)-8 \sin (2 x)+12 x}{32}+C
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: Find the value of $\int \sin ^2 x \cos ^2 x d x$

1) $\frac{x}{2}-\frac{\sin 2 x}{2}+C$
2) $\frac{x}{4}-\frac{\sin 4 x}{4}+C$
3) $\frac{1}{8}\left(x-\frac{1}{4} \sin (4 x)\right)+C$
4) $\frac{x}{2}-\frac{\sin 2 x}{4}+C$

Solution

$=\int \sin ^2 x \cos ^2 x d x=\int \frac{\sin ^2(2 x)}{4} d x=\int \frac{1-\cos (4 x)}{8} d x=\frac{1}{8}\left(x-\frac{1}{4} \sin (4 x)\right)+C$

Hence, the answer is the option (3).

Example 3: $\int\left(\tan ^2 x+\cot ^2 x\right) d x$

$\begin{aligned} & \text { 1) } \tan x+\cot x+C \\ & \text { 2) } \tan x+\cot x-2+C \\ & \text { 3) }-2 x+\tan (x)-\cot (x)+C \\ & \text { 4) } \sec x+\csc x+C\end{aligned}$

Solution

$\begin{aligned} & I=\int \tan ^2(x) d x+\int \cot ^2(x) d x \\ & \int \tan ^2(x) d x=\int-1+\sec ^2(x) d x=-x+\tan (x) \\ & \int \cot ^2(x) d x=\int-1+\csc ^2(x) d x=-x-\cot (x) \\ & I=-2 x+\tan (x)-\cot (x)+C\end{aligned}$

Hence, the answer is the option (3).

Example 4: Integrate $\int \tan ^4 x d x$

1) $\frac{\tan ^3 x}{3}-\tan x+C$
2) $\frac{\tan ^3 x}{3}-\tan x+x+C$
3) $\frac{\tan ^3 x}{3}+C$
4) $\frac{\tan ^3 x}{3}+\tan x+C$

Solution

$\begin{aligned} & \int \tan ^4 x d x \\ & =\int \tan ^2 x\left(\sec ^2 x-1\right) d x \\ & =\int \tan ^2 x \sec ^2 x d x-\int \tan ^2 x d x \\ & =\int \tan ^2 x \sec ^2 x d x-\int\left(\sec ^2 x-1\right) d x \\ & =\frac{\tan ^3 x}{3}-\tan x+x+C\end{aligned}$

Hence, the answer is the option (2).

Example 5: Integrate $\int \cot ^5 x d x$

1) $\frac{\cot ^4 x}{4}-\frac{\cot ^2 x}{2}+C$

2) $-\ln (\csc (x))-\frac{\csc ^4(x)}{4}+\csc ^2(x)+c$

3) $\frac{\cot ^4 x}{4}-\cot ^2 x+\ln \sin x+C$

4)None of these

Solution

$\begin{aligned} & \int \cot ^5 x d x=\int \cot x\left(\csc ^2 x-1\right)^2 d x \\ & u=\csc (x): \mathrm{d} x=-\frac{1}{\cot (x) \csc (x)} \mathrm{d} u \\ & =-\int \frac{\left(u^2-1\right)^2}{u} \mathrm{~d} u \\ & =-\ln (\csc (x))-\frac{\csc ^4(x)}{4}+\csc ^2(x)+c\end{aligned}$

Hence, the answer is the option (2).