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 \[\int{\sin x.dx}=-\cos x+C\]
\[\int{\cos x.dx}=\sin x+C\]
\[\begin{align}
  & \therefore {{I}_{n}}=-\cos x.{{\sin }^{n-1}}x-\int{\left( n-1 \right){{\sin }^{n-2}}x.\cos x\left( -\cos x \right)dx} \\
 & {{I}_{n}}=-\cos x.{{\sin }^{n-1}}x+\left( n-1 \right)\int{{{\sin }^{n-2}}x{{\cos }^{2}}x.dx} \\
\end{align}\]
In the above equation, \[{{\cos }^{2}}x=1-{{\sin }^{2}}x\]
\[\begin{align}
  & \left( n-1 \right)\left[ {{\sin }^{n-2}}x\left( 1-{{\sin }^{2}}x \right) \right]=\left( n-1 \right){{\sin }^{n-2}}x-\left( n-1 \right){{\sin }^{\left( n-2+2 \right)}}x \\
 & \left( n-1 \right)\left[ {{\sin }^{n-2}}x\left( 1-{{\sin }^{2}}x \right) \right]=\left( n-1 \right){{\sin }^{n-2}}x-\left( n-1 \right){{\sin }^{n}}x \\
 & \Rightarrow {{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right)\int{{{\sin }^{n-2}}xdx-\left( n-1 \right)\int{{{\sin }^{n}}x.dx}} \\
 & \therefore {{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}}-\left( n-1 \right){{I}_{n}} \\
\end{align}\]
Here, \[{{I}_{n}}={{\sin }^{n}}xdx\]
\[{{I}_{n-2}}={{\sin }^{n-2}}xdx\]
\[\begin{align}
  & \Rightarrow {{I}_{n}}+\left( n-1 \right){{I}_{n}}=-\cos x.{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}} \\
 & \therefore n{{I}_{n}}=-\cos x{{\sin }^{n-1}}x+\left( n-1 \right){{I}_{n-2}} \\
 & \therefore {{I}_{n}}=\dfrac{-\cos x{{\sin }^{n-1}}x}{n}+\dfrac{\left( n-1 \right)}{n}{{I}_{n-2}} \\
 & \Rightarrow {{I}_{n}}=\dfrac{1}{n}\left[ \left( n-1 \right){{I}_{n-2}}-\cos x.{{\sin }^{n-1}}x \right] \\
\end{align}\]
Now, putting the interval \[\left( 0,\dfrac{\pi }{2} \right)\]
\[\begin{align}
  & {{I}_{n}}=\left[ \dfrac{1}{n}\left[ \left( n-1 \right){{I}_{n-2}}-\cos x.{{\sin }^{n-1}}x \right] \right]_{0}^{\dfrac{\pi }{2}} \\
 & \Rightarrow \dfrac{1}{n}\left[ \left( n-1 \right){{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-\cos \left( \dfrac{\pi }{2} \right){{\sin }^{n-1}}\left( \dfrac{\pi }{2} \right) \right]-\dfrac{1}{n}\left[ 0+0 \right] \\
 & =\dfrac{1}{n}\left[ \left[ n-1 \right]{{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-\cos \left( \dfrac{\pi }{2} \right){{\sin }^{n-1}}\left( \dfrac{\pi }{2} \right) \right] \\
 & \because \cos \left( \dfrac{\pi }{2} \right)=0,\sin \left( \dfrac{\pi }{2} \right)=1 \\
 & =\dfrac{1}{n}\left[ \left( n-1 \right){{\sin }^{n-2}}\left( \dfrac{\pi }{2} \right)-0 \right]=\dfrac{1}{n}\left( n-1 \right) \\
 & \therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{n}}xdx}=\dfrac{n-1}{n} \\
\end{align}\]

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 \[\dfrac{\left( n-1 \right)}{n}\].