Question

Find the value of integral \[\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\sin 2x.\sin 3x.\sin 4x.dx} \]
(a) \[\dfrac{\pi }{4}\]
(b) \[\dfrac{\pi }{8}\]
(c) \[\dfrac{\pi }{16}\]
(d) \[\dfrac{\pi }{32}\]

Answer 1

Hint: We solve this problem by using the simple formulas of trigonometry.
First, we regroup the terms in the integral in such a way that the odd times of \['x'\] at one side and the even times of \['x'\] at one side.
Then we use the formula of composite angles of sine ratio that is
\[2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)\]
Also we use the formula of composite angle of cosine ratio that is
\[2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\]
We also have the formula of half angle for cosine ratio that is
\[{{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}\]
After applying the required formulas we get the integral as the individual terms of cosine ratio then we use the direct formula of definite integration that is
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\cos nx.dx}=\left\{ \begin{align}
  & 0,\text{ when }n\text{ is even} \\
 & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\
\end{align} \right.\]

Complete step-by-step solution
We are given that the integral as \[\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\sin 2x.\sin 3x.\sin 4x.dx}\]
Let us assume that the given integral as
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\sin 2x.\sin 3x.\sin 4x.dx}\]
Now, let us regroup the terms in such a way that the odd times of \['x'\] at one side and the even times of \['x'\] at one side then we get
\[\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin x.\sin 3x \right)\left( \sin 2x.\sin 4x \right)dx}\]
We know that the formula of composite angle of sine ratio that is
\[2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\cos \left( 3x-x \right)-\cos \left( 3x+x \right)}{2} \right)\left( \dfrac{\cos \left( 4x-2x \right)-\cos \left( 4x+2x \right)}{2} \right)dx} \\
 & \Rightarrow I=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cos 2x-\cos 4x \right)\left( \cos 2x-\cos 6x \right)dx} \\
\end{align}\]
Now, by multiplying the terms in above equation we get
\[\Rightarrow I=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\cos }^{2}}2x-\cos 2x\cos 6x-\cos 2x\cos 4x+\cos 4x\cos 6x \right)dx}.....equation(i)\]
We know that the half angle formula of cosine ratio that is
\[{{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2}\]
We also know that the composite angle formula for the cosine ratio that is
\[2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\]
Now, by using the above two formulas to equation (i) we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1+\cos 4x}{2}-\dfrac{\cos 8x}{2}-\dfrac{\cos 4x}{2}-\dfrac{\cos 6x}{2}-\dfrac{\cos 2x}{2}+\dfrac{\cos 10x}{2}+\dfrac{\cos 2x}{2} \right)dx} \\
 & \Rightarrow I=\dfrac{1}{4}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{1}{2}+\dfrac{\cos 4x}{2}-\dfrac{\cos 8x}{2}-\dfrac{\cos 4x}{2}-\dfrac{\cos 6x}{2}+\dfrac{\cos 10x}{2} \right)dx} \\
\end{align}\]
Now, let us separate the integral for each and every term in the above equation we get
\[\Rightarrow I=\dfrac{1}{4}\left[ \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1}{2}.dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos 4x}{2}.dx}-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 8x.dx}-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 4x.dx}-\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 6x.dx}+\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\cos 10x.dx} \right]\]
We know that the standard result of definite integrals that is
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\cos nx.dx}=\left\{ \begin{align}
  & 0,\text{ when }n\text{ is even} \\
 & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\
\end{align} \right.\]
\[\int\limits_{a}^{b}{dx}=b-a\]
By using this formula to above equation we get
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{4}\left[ \dfrac{1}{2}\left( \dfrac{\pi }{2}-0 \right)+\left( \dfrac{1}{2}\times 0 \right)-\left( \dfrac{1}{2}\times 0 \right)-\left( \dfrac{1}{2}\times 0 \right)-\left( \dfrac{1}{2}\times 0 \right)+\left( \dfrac{1}{2}\times 0 \right) \right] \\
 & \Rightarrow I=\dfrac{\pi }{16} \\
\end{align}\]
Therefore the value of given integral is
\[\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\sin 2x.\sin 3x.\sin 4x.dx}=\dfrac{\pi }{16}\]
So, option (c) is the correct answer.

Note: Here we used the direct result for the definite integral that is
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\cos nx.dx}=\left\{ \begin{align}
  & 0,\text{ when }n\text{ is even} \\
 & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\
\end{align} \right.\]
If we cannot remember this we can prove this by using the integration.
Let us assume that
\[A=\int\limits_{0}^{\dfrac{\pi }{2}}{\cos nx.dx}\]
We know that the formula for the integration that is
\[\int{\cos nx.dx}=\dfrac{1}{n}\sin nx\]
So, by using this formula we get
\[\Rightarrow A=\dfrac{1}{n}\left[ \sin nx \right]_{0}^{\dfrac{\pi }{2}}\]
Now, by applying the limits we get
\[\begin{align}
  & \Rightarrow A=\dfrac{1}{n}\sin \dfrac{n\pi }{2}-\dfrac{1}{n}\sin 0 \\
 & \Rightarrow A=\dfrac{1}{n}\sin \dfrac{n\pi }{2} \\
\end{align}\]
Here we can see if \['n'\] is even we can say that the value of \[\dfrac{n\pi }{2}\] will be the multiple of \[\pi \]
We know that the sine ratio for integral multiples of \[\pi \] will be 0 that is
\[\Rightarrow \sin k\pi =0\] where, \['k'\] is an integer.
But we cannot say anything when \['n'\] is odd.
Therefore we can write that
\[\int\limits_{0}^{\dfrac{\pi }{2}}{\cos nx.dx}=\left\{ \begin{align}
  & 0,\text{ when }n\text{ is even} \\
 & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\
\end{align} \right.\]
We can directly use this in the problems as we did in this question.