Definite IntegrationmediumFree

∫ sin⁶x/((sin⁶x+cos⁶x)(1+e⁻ˣ)) dx, −2π to 2π | JEE

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

Question

\displaystyle\int_{-2\pi}^{2\pi}\dfrac{\sin^{6}x}{\big(\sin^{6}x+\cos^{6}x\big)\big(1+e^{-x}\big)}\,dx

equals

2\pi

\pi

correct

\dfrac{\pi}{2}

4\pi

Solution

Step 1: Isolate the even part of the integrand. Let

\phi(x)=\dfrac{\sin^{6}x}{\sin^{6}x+\cos^{6}x}

, which is even, and write

I=\int_{-2\pi}^{2\pi}\dfrac{\phi(x)}{1+e^{-x}}\,dx.

Step 2: Replace

x\mapsto-x

. Since

\phi

is even and the limits are symmetric,

I=\int_{-2\pi}^{2\pi}\dfrac{\phi(x)}{1+e^{x}}\,dx.

Add this to the original form of

I

2I=\int_{-2\pi}^{2\pi}\phi(x)\left(\dfrac{1}{1+e^{-x}}+\dfrac{1}{1+e^{x}}\right)dx.

Step 3: The bracket simplifies. Since

\dfrac{1}{1+e^{-x}}+\dfrac{1}{1+e^{x}}=\dfrac{e^{x}}{e^{x}+1}+\dfrac{1}{1+e^{x}}=1

2I=\int_{-2\pi}^{2\pi}\phi(x)\,dx=2\int_{0}^{2\pi}\dfrac{\sin^{6}x}{\sin^{6}x+\cos^{6}x}\,dx,

using that

\phi

is even. Step 4: Exploit the

\sin\leftrightarrow\cos

symmetry. Over

[0,2\pi]

the integrals of

\dfrac{\sin^{6}x}{\sin^{6}x+\cos^{6}x}

and

\dfrac{\cos^{6}x}{\sin^{6}x+\cos^{6}x}

are equal, and they add to

\displaystyle\int_{0}^{2\pi}1\,dx=2\pi

. Hence each equals

\pi

2I=2\pi\ \Rightarrow\ I=\pi.

Correct answer: (2)

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