Question

Obtain the reduction formula for $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{n}}xdx}$ for an integer $n\ge 2.$

Answer 1

Hint: Solve the integral by putting ${{\sin }^{n}}x={{\sin }^{n-1}}x.\sin x$ and expand it using $\int{uv=u\int{v.dx-\int{\dfrac{d}{dx}\left( u \right)}\int{\left( vdx \right)dx}}}$
(Use the integral formula to solve the problem.)

Complete step-by-step answer:
Given, $I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{n}}xdx}$.
Let, ${{I}_{n}}=\int{{{\sin }^{n}}xdx}$
Let us write ${{\sin }^{n}}x={{\sin }^{n-1}}x.\sin xdx$
$\Rightarrow {{I}_{n}}=\int{\left( {{\sin }^{n-1}}x \right)\left( \sin x \right)dx}$
Let us take $u={{\sin }^{n-1}}x$and $v=\sin x$
$\therefore {{I}_{n}}=\int{uv}$
i.e. ${{I}_{n}}=u\int{v.dx}-\int{\dfrac{d}{dx}\left( u \right)\int{\left( v.dx \right)dx}}$
$\therefore {{I}_{n}}={{\sin }^{n-1}}x\int{\sin x.dx}-\int{\dfrac{d}{dx}}\left( {{\sin }^{n-1}}x \right)\int{\left( \sin x.dx \right)dx}$
$\Rightarrow$We know $\int{\sin x.dx}=-\cos x+C$
$\int{\cos x.dx}=\sin x+C$

& \therefore {{I}_{n}}=-\cos x.{{\sin }^{n-1}}x-\int{\left( n-1 \right){{\sin }^{n-2}}x.\cos x\left( -\cos x \right)dx} \\\ & {{I}_{n}}=-\cos x.{{\sin }^{n-1}}x+\left( n-1 \right)\int{{{\sin }^{n-2}}x{{\cos }^{2}}x.dx} \\\ \end{aligned}$$ In the above equation, $${{\cos }^{2}}x=1-{{\sin }^{2}}x$$ $$\begin{aligned} & \left( n-1 \right)\left[ {{\sin }^{n-2}}x\left( 1-{{\sin }^{2}}x \right) \right]=\left( n-1 \right){{\sin }^{n-2}}x-\left( n-1 \right){{\sin }^{\left( n-2+2 \right)}}x \\\ & \left( n-1 \right)\left[ {{\sin }^{n-2}}x\left( 1-{{\sin }^{2}}x \right) \right]=\left( n-1 \right){{\sin }^{n-2}}x-\left( n-1 \right){{\sin }^{n}}x \\\ & \Rightarrow {{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right)\int{{{\sin }^{n-2}}xdx-\left( n-1 \right)\int{{{\sin }^{n}}x.dx}} \\\ & \therefore {{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}}-\left( n-1 \right){{I}_{n}} \\\ \end{aligned}$$ Here, $${{I}_{n}}={{\sin }^{n}}xdx$$ $${{I}_{n-2}}={{\sin }^{n-2}}xdx$$ $$\begin{aligned} & \Rightarrow {{I}_{n}}+\left( n-1 \right){{I}_{n}}=-\cos x.{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}} \\\ & \therefore n{{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}} \\\ & \therefore {{I}_{n}}=\dfrac{-\cos x{{\sin }^{n-1}}x}{n}+\dfrac{\left( n-1 \right)}{n}{{I}_{n-2}} \\\ & \Rightarrow {{I}_{n}}=\dfrac{1}{n}\left[ \left( n-1 \right){{I}_{n-2}}-\cos x.{{\sin }^{n-1}}x \right] \\\ \end{aligned}$$ Now, putting the interval $$\left( 0,\dfrac{\pi }{2} \right)$$ $$\begin{aligned} & {{I}_{n}}=\left[ \dfrac{1}{n}\left[ \left( n-1 \right){{I}_{n-2}}-\cos x.{{\sin }^{n-1}}x \right] \right]_{0}^{\dfrac{\pi }{2}} \\\ & \Rightarrow \dfrac{1}{n}\left[ \left( n-1 \right){{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-\cos \left( \dfrac{\pi }{2} \right){{\sin }^{n-1}}\left( \dfrac{\pi }{2} \right) \right]-\dfrac{1}{n}\left[ 0+0 \right] \\\ & =\dfrac{1}{n}\left[ \left[ n-1 \right]{{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-\cos \left( \dfrac{\pi }{2} \right){{\sin }^{n-1}}\left( \dfrac{\pi }{2} \right) \right] \\\ & \because \cos \left( \dfrac{\pi }{2} \right)=0,\sin \left( \dfrac{\pi }{2} \right)=1 \\\ & =\dfrac{1}{n}\left[ \left( n-1 \right){{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-0 \right]=\dfrac{1}{n}\left( n-1 \right) \\\ & \therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{n}}xdx}=\dfrac{n-1}{n} \\\ \end{aligned}$$ Note: After solving integral by putting $${{\sin }^{n}}x={{\sin }^{n-1}}x.\sin x$$ and once the reduction is obtained remember to put $$\left( 0,\dfrac{\pi }{2} \right)$$ to obtain $$\dfrac{\left( n-1 \right)}{n}$$.