The value of the integral \int\limits_ - \pi \over 2^\pi \over 2 \sin ^4 x( 1 + \log ( 2 + \sin x \over 2 - \sin x ) )dx is :

Definite Integration PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q41 with Solution

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Correct answer: C

Let $I=\int_{-\pi/2}^{\pi/2} \sin^4 x\left(1+\log\frac{2+\sin x}{2-\sin x}\right)\,dx.$ We split it as $I=\int_{-\pi/2}^{\pi/2}\sin^4 x\,dx+\int_{-\pi/2}^{\pi/2}\sin^4 x\log\frac{2+\sin x}{2-\sin x}\,dx.$ So write $I=I_1+I_2.$
Evaluate the parity of the second integrand.

Define $f(x)=\sin^4 x\log\frac{2+\sin x}{2-\sin x}.$ Now, $f(-x)=\sin^4(-x)\log\frac{2+\sin(-x)}{2-\sin(-x)}.$ Since $\sin(-x)=-\sin x$ and $\sin^4(-x)=\sin^4 x$ ,

=-\sin^4 x\log\frac{2+\sin x}{2-\sin x}=-f(x).$$ Hence $f(x)$ is an odd function. Therefore, over the symmetric interval $\left[-\frac\pi2,\frac\pi2\right]$, $$I_2=\int_{-\pi/2}^{\pi/2} f(x)\,dx=0.$$ So, $$I=I_1=\int_{-\pi/2}^{\pi/2}\sin^4 x\,dx.$$ 3. Now compute $I_1$. Using symmetry, $$I_1=2\int_0^{\pi/2}\sin^4 x\,dx.$$ Use the standard identity $$\sin^4 x=\left(\sin^2 x\right)^2=\left(\frac{1-\cos 2x}{2}\right)^2 =\frac{3}{8}-\frac12\cos 2x+\frac18\cos 4x.$$ Thus, $$I_1=2\int_0^{\pi/2}\left(\frac38-\frac12\cos 2x+\frac18\cos 4x\right)dx.$$ Integrating, $$I_1=2\left[\frac38 x-\frac14\sin 2x+\frac1{32}\sin 4x\right]_0^{\pi/2}. $$ At $x=\frac\pi2$, both sine terms vanish; at $x=0$, they also vanish. Hence $$I_1=2\left(\frac38\cdot\frac\pi2\right)=\frac{3\pi}{8}. $$ 4. Therefore, $$\boxed{I=\frac{3\pi}{8}}.$$ 5. Option check: - A: $0$ — incorrect - B: $\frac34$ — incorrect - C: $\frac{3\pi}{8}$ — correct - D: $\frac{3\pi}{16}$ — incorrect

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View written solutionFree

Correct answer: C

Let $I=\int_{-\pi/2}^{\pi/2} \sin^4 x\left(1+\log\frac{2+\sin x}{2-\sin x}\right)\,dx.$ We split it as $I=\int_{-\pi/2}^{\pi/2}\sin^4 x\,dx+\int_{-\pi/2}^{\pi/2}\sin^4 x\log\frac{2+\sin x}{2-\sin x}\,dx.$ So write $I=I_1+I_2.$
Evaluate the parity of the second integrand.

Define $f(x)=\sin^4 x\log\frac{2+\sin x}{2-\sin x}.$ Now, $f(-x)=\sin^4(-x)\log\frac{2+\sin(-x)}{2-\sin(-x)}.$ Since $\sin(-x)=-\sin x$ and $\sin^4(-x)=\sin^4 x$ ,

=-\sin^4 x\log\frac{2+\sin x}{2-\sin x}=-f(x).$$ Hence $f(x)$ is an odd function. Therefore, over the symmetric interval $\left[-\frac\pi2,\frac\pi2\right]$, $$I_2=\int_{-\pi/2}^{\pi/2} f(x)\,dx=0.$$ So, $$I=I_1=\int_{-\pi/2}^{\pi/2}\sin^4 x\,dx.$$ 3. Now compute $I_1$. Using symmetry, $$I_1=2\int_0^{\pi/2}\sin^4 x\,dx.$$ Use the standard identity $$\sin^4 x=\left(\sin^2 x\right)^2=\left(\frac{1-\cos 2x}{2}\right)^2 =\frac{3}{8}-\frac12\cos 2x+\frac18\cos 4x.$$ Thus, $$I_1=2\int_0^{\pi/2}\left(\frac38-\frac12\cos 2x+\frac18\cos 4x\right)dx.$$ Integrating, $$I_1=2\left[\frac38 x-\frac14\sin 2x+\frac1{32}\sin 4x\right]_0^{\pi/2}. $$ At $x=\frac\pi2$, both sine terms vanish; at $x=0$, they also vanish. Hence $$I_1=2\left(\frac38\cdot\frac\pi2\right)=\frac{3\pi}{8}. $$ 4. Therefore, $$\boxed{I=\frac{3\pi}{8}}.$$ 5. Option check: - A: $0$ — incorrect - B: $\frac34$ — incorrect - C: $\frac{3\pi}{8}$ — correct - D: $\frac{3\pi}{16}$ — incorrect

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