The value of
Question:

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is

  1. $\frac{\pi}{4}$

  2. $4 \pi$

  3. $\frac{\pi}{2}$

  4. $2 \pi$


Correct Option: 1

Solution:

$I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ (using king)

$I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{-x}} d x=\int_{-\pi / 2}^{\pi / 2} \frac{3^{x} \cos ^{2} x}{1+3^{x}} d x$

$2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2} \frac{\left(1+3^{\mathrm{x}}\right) \cos ^{2} \mathrm{x}}{1+3^{\mathrm{x}}} \mathrm{dx}$

$=\int_{-\pi / 2}^{\pi / 2} \cos ^{2} x d x=2 \int_{0}^{\pi / 2} \cos ^{2} x d x$

$\Rightarrow I=\int_{0}^{\pi / 2} \cos ^{2} x d x=\frac{\pi}{4}$

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