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∫ sinx sin2x sin3x sin4x dx, 0 to π/2 | JEE

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\int_{0}^{\pi/2}\sin x\,\sin 2x\,\sin 3x\,\sin 4x\,dx

equals

\dfrac{\pi}{4}

\dfrac{\pi}{8}

\dfrac{\pi}{16}

correct

\dfrac{\pi}{32}

Solution

Step 1: Call the integral

I

and apply the property

\displaystyle\int_{0}^{\pi/2}f(x)\,dx=\int_{0}^{\pi/2}f\!\left(\dfrac{\pi}{2}-x\right)dx

. Under

x\mapsto\dfrac{\pi}{2}-x

\sin x\to\cos x,\quad \sin 2x\to\sin 2x,\quad \sin 3x\to\cos 3x,\quad \sin 4x\to-\sin 4x.

But

\sin 4x\to\sin\!\big(2\pi-4x\big)=-\sin 4x

; combined with the sign bookkeeping the reflected integrand is

I=\int_{0}^{\pi/2}\cos x\,\sin 2x\,\cos 3x\,\sin 4x\,dx.

Step 2: Add the two forms of

I

2I=\int_{0}^{\pi/2}\sin 2x\,\sin 4x\,\big(\sin x\sin 3x+\cos x\cos 3x\big)\,dx=\int_{0}^{\pi/2}\sin 2x\,\sin 4x\,\cos 2x\,dx,

using

\cos x\cos 3x+\sin x\sin 3x=\cos(3x-x)=\cos 2x

. Step 3: Combine

\sin 2x\cos 2x=\dfrac12\sin 4x

, so

2I=\dfrac12\int_{0}^{\pi/2}\sin^{2}4x\,dx=\dfrac12\int_{0}^{\pi/2}\dfrac{1-\cos 8x}{2}\,dx.

Step 4: Evaluate. Over

[0,\pi/2]

\displaystyle\int_{0}^{\pi/2}\cos 8x\,dx=0

, hence

2I=\dfrac14\cdot\dfrac{\pi}{2}=\dfrac{\pi}{8}\ \Rightarrow\ I=\dfrac{\pi}{16}.

Correct answer: (3)

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