The value of \int\limits_ - \pi /2^\pi /2 \cos ^2x \over 1 + 3^x dx is :

Definite Integration PYQ — JEE Main 2021 · 26 Feb · Shift 1 · Q26 with Solution

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Correct answer: D

Let $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x}\,dx.$ We use the standard symmetry property for integrals over $[-a,a]$ .
Consider $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{1+3^{-x}}\,dx$ because replacing $x\to -x$ does not change the limits after simplification. Since $\cos^2(-x)=\cos^2 x$ , $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^{-x}}\,dx.$ Now, $\frac{1}{1+3^{-x}}=\frac{3^x}{1+3^x}.$ So, $I=\int_{-\pi/2}^{\pi/2} \frac{3^x\cos^2 x}{1+3^x}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi/2}^{\pi/2} \cos^2 x\left(\frac{1}{1+3^x}+\frac{3^x}{1+3^x}\right)dx.$ The bracket simplifies to $1$ , hence $2I=\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx.$ Therefore, $I=\frac12\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx.$
Now evaluate the remaining integral: $\cos^2 x=\frac{1+\cos 2x}{2}.$ Thus,

=\int_{-\pi/2}^{\pi/2} \frac{1+\cos 2x}{2}\,dx =\frac12\left[x+\frac{\sin 2x}{2}\right]_{-\pi/2}^{\pi/2}.$$ Since $\sin(\pi)=\sin(-\pi)=0$, $$\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx=\frac12\left(\frac\pi2-\left(-\frac\pi2\right)\right)=\frac\pi2.$$ Hence, $$I=\frac12\cdot \frac\pi2=\frac\pi4.$$ 5. Therefore the correct option is $$\boxed{\text{D }\frac\pi4}.$$ 6. Comparison with stored answer: Stored correct answer is D, which matches our result.

View written solutionFree

Correct answer: D

Let $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^x}\,dx.$ We use the standard symmetry property for integrals over $[-a,a]$ .
Consider $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2(-x)}{1+3^{-x}}\,dx$ because replacing $x\to -x$ does not change the limits after simplification. Since $\cos^2(-x)=\cos^2 x$ , $I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1+3^{-x}}\,dx.$ Now, $\frac{1}{1+3^{-x}}=\frac{3^x}{1+3^x}.$ So, $I=\int_{-\pi/2}^{\pi/2} \frac{3^x\cos^2 x}{1+3^x}\,dx.$
Add the two expressions for $I$ : $2I=\int_{-\pi/2}^{\pi/2} \cos^2 x\left(\frac{1}{1+3^x}+\frac{3^x}{1+3^x}\right)dx.$ The bracket simplifies to $1$ , hence $2I=\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx.$ Therefore, $I=\frac12\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx.$
Now evaluate the remaining integral: $\cos^2 x=\frac{1+\cos 2x}{2}.$ Thus,

=\int_{-\pi/2}^{\pi/2} \frac{1+\cos 2x}{2}\,dx =\frac12\left[x+\frac{\sin 2x}{2}\right]_{-\pi/2}^{\pi/2}.$$ Since $\sin(\pi)=\sin(-\pi)=0$, $$\int_{-\pi/2}^{\pi/2} \cos^2 x\,dx=\frac12\left(\frac\pi2-\left(-\frac\pi2\right)\right)=\frac\pi2.$$ Hence, $$I=\frac12\cdot \frac\pi2=\frac\pi4.$$ 5. Therefore the correct option is $$\boxed{\text{D }\frac\pi4}.$$ 6. Comparison with stored answer: Stored correct answer is D, which matches our result.