The value of \int\limits_ - \pi /2^\pi /2 \sin ^2x \over 1 + 2^x dx is

Definite Integration PYQ — JEE Main 2018 · Shift 0 · Q40 with Solution

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Correct answer: A

We use the standard symmetry trick for integrals over $[-a,a]$ .

Substitute $x\mapsto -x$ : $I=\int_{-\pi/2}^{\pi/2} \frac{\sin^2(-x)}{1+2^{-x}}\,dx.$ Since $\sin^2(-x)=\sin^2 x$ , $I=\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^{-x}}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi/2}^{\pi/2} \sin^2 x\left(\frac{1}{1+2^x}+\frac{1}{1+2^{-x}}\right)dx.$ Now simplify the bracket: $\frac{1}{1+2^{-x}}=\frac{2^x}{1+2^x}.$ Hence, $\frac{1}{1+2^x}+\frac{1}{1+2^{-x}}=\frac{1}{1+2^x}+\frac{2^x}{1+2^x}=1.$ Therefore, $2I=\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx.$ So, $I=\frac12\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx.$
Evaluate the remaining integral: Using $\sin^2 x=\frac{1-\cos 2x}{2},$ we get

=\int_{-\pi/2}^{\pi/2} \frac{1-\cos 2x}{2}\,dx.$$ But over one symmetric interval of length $\pi$, $$\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx=\frac{\pi}{2}.$$ Hence, $$I=\frac12\cdot \frac{\pi}{2}=\frac{\pi}{4}.$$ 5. Check options: - A: $\dfrac{\pi}{4}$ ✅ - B: $\dfrac{\pi}{8}$ ❌ - C: $\dfrac{\pi}{2}$ ❌ - D: $4\pi$ ❌ Therefore, the correct answer is **A**.

View written solutionFree

Correct answer: A

We use the standard symmetry trick for integrals over $[-a,a]$ .

Substitute $x\mapsto -x$ : $I=\int_{-\pi/2}^{\pi/2} \frac{\sin^2(-x)}{1+2^{-x}}\,dx.$ Since $\sin^2(-x)=\sin^2 x$ , $I=\int_{-\pi/2}^{\pi/2} \frac{\sin^2 x}{1+2^{-x}}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi/2}^{\pi/2} \sin^2 x\left(\frac{1}{1+2^x}+\frac{1}{1+2^{-x}}\right)dx.$ Now simplify the bracket: $\frac{1}{1+2^{-x}}=\frac{2^x}{1+2^x}.$ Hence, $\frac{1}{1+2^x}+\frac{1}{1+2^{-x}}=\frac{1}{1+2^x}+\frac{2^x}{1+2^x}=1.$ Therefore, $2I=\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx.$ So, $I=\frac12\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx.$
Evaluate the remaining integral: Using $\sin^2 x=\frac{1-\cos 2x}{2},$ we get

=\int_{-\pi/2}^{\pi/2} \frac{1-\cos 2x}{2}\,dx.$$ But over one symmetric interval of length $\pi$, $$\int_{-\pi/2}^{\pi/2} \sin^2 x\,dx=\frac{\pi}{2}.$$ Hence, $$I=\frac12\cdot \frac{\pi}{2}=\frac{\pi}{4}.$$ 5. Check options: - A: $\dfrac{\pi}{4}$ ✅ - B: $\dfrac{\pi}{8}$ ❌ - C: $\dfrac{\pi}{2}$ ❌ - D: $4\pi$ ❌ Therefore, the correct answer is **A**.