Reduction Formula - Practice Questions & MCQ

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

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
\end{aligned}
$
$
\begin{array}{rlrl}
\therefore & n I_n & =-\sin ^{n-1} x \cos x+(n-1) I_{n-2} \\
\Rightarrow & & I_n & =-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}
\end{array}
$

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) $\int \tan ^{\mathrm{n}} \mathrm{xdx}$

Let $\quad I_n=\int \tan ^n x d x$
$
\begin{aligned}
\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 \quad 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}
$

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

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

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

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
$
\left.\begin{array}{l}
\quad \begin{array}{l}
\quad \\
\sec ^{n-2} x \tan x-\int(n-2) \sec ^{n-3} x \sec x \tan x \cdot \tan x d x \\
\\
= \\
\sec ^{n-2} x \tan x-(n-2) \int \sec ^{n-2} x\left(\sec ^2 x-1\right) d x \\
\\
= \\
\sec ^{n-2} x \tan x-(n-2) I_n+(n-2) I_{n-2}
\end{array} \\
\Rightarrow(n-1) I_n=\sec ^{n-2} x \tan x+(n-2) I_{n-2}
\end{array}\right\}
$

(f) $\int \csc ^{\mathrm{n}} \mathrm{x} \mathrm{dx}$

Let $I_n=\int \csc ^n x d x=\int \csc ^{n-2} x \csc ^2 x d x$ $=\csc ^{n-2} x(-\cot x)-\int(n-2) \csc ^{n-2} x\left(\csc ^2 x-1\right) d x$ $=-\csc ^{n-2} x(-\cot x)-(n-2) \int\left(\csc ^n x-\csc ^{n-2} x\right) d x$ $=-\csc ^{n-2} x \cot x-(n-2) I_n+(n-2) I_{n-2}$
$\therefore \quad(n-1) I_n=-\csc ^{n-2} x \cot x+(n-2) I_{n-2}$
or $\quad I_n=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} I_{n-2}$
$\therefore \int \csc ^n x d x=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \int \csc ^{n-2} x d x$

Integration of the type $\int \cos ^{\mathrm{m}} \mathrm{x} \sin \mathrm{nx} \mathbf{d x}$
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 first function and $\sin \mathrm{nx}$ as second function.
$
\left.\begin{array}{rl}
\quad & =-\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
\end{array}\right] \begin{aligned}
{[\mathrm{using} \sin (n-1) x=\sin n x \cos x-\cos n x \sin x}
\end{aligned} \quad \begin{aligned}
\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 a 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. $\int \sin ^m x \cos n x d x=\frac{n \sin ^m x \sin n x}{m^2-n^2}-\frac{m \sin ^{m-1} x \cos x \cos n x}{m^2-n^2}+\frac{m(m-1)}{m^2-n^2} \int \sin ^{m-2} x \cos n x d x$