Q51 / 90 · KVPY 2012 (SB/SX) · Mathematics
The value of the integral $\frac{\pi}{-\pi} \frac{\cos ^{2} x}{1+\mathrm{a}^{\mathrm{x}}} \mathrm{d} \mathrm{x}$, where $\mathrm{a}>0$, is -
- A$\pi$
- B$\mathrm{a} \pi$
- Correct answerC$\pi / 2$
- D$2 \pi$
Solution
$$
\begin{array}{l}
I=\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x \ldots \ldots \ldots \ldots \ldots \ldots(1) \quad \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \\
I=\int_{-\pi}^{\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d x \\
I=\int_{-\pi}^{\pi} \frac{a^{x} \cos ^{2} x}{1+a^{x}} d x \ldots \ldots \ldots \ldots \ldots \ldots . . (2)
\end{array}
$$
add equation (1) and (2)
$2 \mathrm{I}=\int_{-\pi}^{\pi} \cos ^{2} \mathrm{x}\left(\frac{1}{1+\mathrm{a}^{\mathrm{x}}}+\frac{\mathrm{a}^{\mathrm{x}}}{1+\mathrm{a}^{\mathrm{x}}}\right) \mathrm{dx}$
$\mathrm{I}=\int_{0}^{\pi} \cos ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=2 \int_{0}^{\pi / 2} \cos ^{2} \mathrm{x} \mathrm{dx}$
$=\pi / 2$
Hence, the correct answer is option: 'C'.