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

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

& 0,\text{ when }n\text{ is even} \\\ & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\\ \end{aligned} \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{aligned} & \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{aligned}$$ Now, by multiplying the terms in above equation we get $$\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{aligned} & \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{aligned}$$ Now, let us separate the integral for each and every term in the above equation we get $$\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{aligned} & 0,\text{ when }n\text{ is even} \\\ & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\\ \end{aligned} \right.$$ $$\int\limits_{a}^{b}{dx}=b-a$$ By using this formula to above equation we get $$\begin{aligned} & \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{aligned}$$ Therefore the value of given integral is $$\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{aligned} & 0,\text{ when }n\text{ is even} \\\ & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\\ \end{aligned} \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{aligned} & \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{aligned}$$ Here we can see if $$'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{aligned} & 0,\text{ when }n\text{ is even} \\\ & \dfrac{1}{n}\sin \left( \dfrac{n\pi }{2} \right),\text{ when }n\text{ is odd} \\\ \end{aligned} \right.$$ We can directly use this in the problems as we did in this question.