Mathematics

The value of the definite integral $$\int _{ 0 }^{ \pi /2 }{ \sin { x } \sin { 2x } \sin { 3x } dx } $$ is equal to:


ANSWER

$$\cfrac{1}{6}$$


SOLUTION
$$\int_0^{\pi/2}\sin x\sin 2x\sin 3x dx$$

$$\Rightarrow \dfrac{1}{2}\int_0^{\pi/2}2\sin x\sin 2x\sin 3x dx$$

$$\Rightarrow \dfrac{1}{2}\int_0^{\pi/2}2\sin x\sin 3x\sin 2x dx$$

We know that       $$2\sin A \sin B=\cos(A-B)-\cos (A+B)$$

$$\Rightarrow \dfrac{1}{2}\int_0^{\pi/2}(\cos (x-3x)-\cos(x+3x))\sin 2x dx$$

$$\Rightarrow \dfrac{1}{2}\int_0^{\pi/2}(\cos 2x-\cos 4x)\sin 2x dx$$

$$\Rightarrow \dfrac{1}{2}\int_0^{\pi/2} \sin 2x \cos 2x dx-\dfrac{1}{2}\int _0^{\pi /2}\sin 2x \cos 4x dx$$

$$\Rightarrow \dfrac{1}{4}\int_0^{\pi/2} 2\sin 2x \cos 2x dx-\dfrac{1}{4}\int _0^{\pi /2}2\sin 2x \cos 4x dx$$

We know that   $$2\sin A \cos B=\sin(A+B)+\sin (A-B)$$

$$\Rightarrow \dfrac{1}{4}\int_0^{\pi/2} (\sin (2x+2x)+\sin (2x-2x))dx-\dfrac{1}{4}\int _0^{\pi /2}(\sin (2x+4x)+\sin (2x-4x)) dx$$

$$\Rightarrow \dfrac{1}{4}\int_0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}(\sin 6x-\sin 2x) dx$$

$$\Rightarrow \dfrac{1}{4}\int_0^{\pi/2} \sin 4xdx-\dfrac{1}{4}\int _0^{\pi /2}\sin 6x dx+\dfrac{1}{4}\int_{0}^{\pi/4}\sin 2x dx$$

$$\Rightarrow \dfrac{1}{4}[\dfrac{\sin 4x}{4}]_0^{\pi/4}-\dfrac{1}{4}[\dfrac{-\sin 6x}{6}]_0^{\pi/4}+\dfrac{1}{4}[\dfrac{-\cos 2x}{2}]_0^{\pi/4}$$

$$\Rightarrow \dfrac{-1}{12}+\dfrac{1}{4}=\dfrac{1}{6}$$
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Single Correct Medium Published on 17th 09, 2020
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