Question

Find the value of $\int\limits_0^{\dfrac{\pi }{2}} {\sin x\sin 2x\sin 3x\sin 4xdx} = $
A. $\dfrac{\pi }{4}$
B. $\dfrac{\pi }{8}$
C. $\dfrac{\pi }{{16}}$
D. \[\dfrac{\pi }{{32}}\]

Answer 1

Hint: We can equate the given integral to I. Then we can apply the property of definite integral that $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} $ and this also will be equal to I. Then we can add the two equations and take the common terms outside. Then we can simplify the remaining terms by using trigonometric identities. Then using identities, we can make the terms inside the integral to the terms that we know the value of the integral. Then we can integrate and apply the limits to get the required answer.

Complete step-by-step answer:
Let $I = \int\limits_0^{\dfrac{\pi }{2}} {\sin x\sin 2x\sin 3x\sin 4xdx} $ … (1)
We know that for a definite integral $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} $ .
Thus $I$ will become,
 \[I = \int\limits_0^{\dfrac{\pi }{2}} {\sin \left( {\dfrac{\pi }{2} - x} \right)\sin \left( {\dfrac{{2\pi }}{2} - 2x} \right)\sin \left( {\dfrac{{3\pi }}{2} - 3x} \right)\sin \left( {\dfrac{{4\pi }}{2} - 4x} \right)dx} \]
On simplification, we get,
 \[I = \int\limits_0^{\dfrac{\pi }{2}} {\sin \left( {\dfrac{\pi }{2} - x} \right)\sin \left( {\pi - 2x} \right)\sin \left( {\dfrac{{3\pi }}{2} - 3x} \right)\sin \left( {2\pi - 4x} \right)dx} \]
We know that $\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$ , $\sin \left( {\pi - 2x} \right) = \sin 2x$ , \[\sin \left( {\dfrac{{3\pi }}{2} - 3x} \right) = - \cos 3x\] , \[\sin \left( {2\pi - 4x} \right) = - \sin 4x\]
Thus, the integral will become,
 \[ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\cos x\sin 2x\cos 3x\sin 4xdx} \] … (2)
Now we can add equations (1) and (2)
 \[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\sin x\sin 2x\sin 3x\sin 4xdx} + \int\limits_0^{\dfrac{\pi }{2}} {\cos x\sin 2x\cos 3x\sin 4xdx} \]
We can take the integral and other common terms from the 2 terms,
 \[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\sin 2x\sin 4xdx\left( {\sin x\sin 3x + \cos x\cos 3x} \right)} \]
We know that $\cos \left( {A - B} \right) = \sin A\sin B + \cos A\cos B$ .
So \[\sin x\sin 3x + \cos x\cos 3x\] will become,
 \[ \Rightarrow \sin x\sin 3x + \cos x\cos 3x = \cos \left( {x - 3x} \right)\]
Hence, we have,
 \[ \Rightarrow \sin x\sin 3x + \cos x\cos 3x = \cos \left( { - 2x} \right)\]
We know that $\cos \left( { - x} \right) = \cos x$ ,
 \[ \Rightarrow \sin x\sin 3x + \cos x\cos 3x = \cos \left( {2x} \right)\]
On applying, this in the sum of the integral, we get,
 \[ \Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{2}} {\sin 2x\sin 4x\cos 2xdx} \]
We know that, $2\sin A\cos A = \sin 2A$ .
So \[\sin 2x\cos 2x = \dfrac{1}{2}\sin 4x\]
 \[ \Rightarrow 2I = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\sin 4x\sin 4xdx} \]
Hence, we have,
 \[ \Rightarrow 2I = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {{{\sin }^2}4xdx} \]
We know that ${\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}$ . On applying this identity, we get,
 \[ \Rightarrow 2I = \dfrac{1}{2}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{1 - \cos 8x}}{2}dx} \]
Taking \[\dfrac{1}{2}\] common we get,
 \[ \Rightarrow 2I = \dfrac{1}{4}\int\limits_0^{\dfrac{\pi }{2}} {1 - \cos 8xdx} \]
We know that $\int {1dx = x} $ and
$\int {\cos ax = \dfrac{{\sin ax}}{a}} $
 \[ \Rightarrow 2I = \dfrac{1}{4}\left[ {x - \dfrac{{\sin 8x}}{8}} \right]_0^{\dfrac{\pi }{2}}\]
On applying the limits, we get,
 \[ \Rightarrow 2I = \dfrac{1}{4}\left[ {\dfrac{\pi }{2} - \dfrac{{\sin 8 \times \dfrac{\pi }{2}}}{8}} \right] - \left[ {0 - \dfrac{{\sin 8 \times 0}}{8}} \right]\]
On simplification we get,
 \[ \Rightarrow 2I = \dfrac{1}{4}\left[ {\dfrac{\pi }{2} - 0} \right] - \left[ {0 - 0} \right]\]
Hence, we have,
 \[ \Rightarrow 2I = \dfrac{1}{4} \times \dfrac{\pi }{2}\]
On dividing throughout with 2, we get,
 \[ \Rightarrow I = \dfrac{\pi }{{16}}\]
Therefore, the required value of the integral is \[\dfrac{\pi }{{16}}\]
So, the correct answer is option C.

Note: The property of definite integrals used in this problem is $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {a + b - x} \right)dx} $ . After applying this, we change sin ratio to cos ratio for odd multiples of $\dfrac{\pi }{2}$ and the sin ratio remains for multiples of $\pi $ . Then the sign of each of the ratios is determined by the quadrant in which the resultant angle lies. The 1st quadrant will have all the ratios positive; 2nd will have only sin as positive; the 3rd quadrant will have only tan as positive and the 4th quadrant have cos as positive.
The trigonometric identities used in this problem are
 $\cos \left( {A - B} \right) = \sin A\sin B + \cos A\cos B$
 $2\sin A\cos A = \sin 2A$
 ${\sin ^2}A = \dfrac{{1 - \cos 2A}}{2}$