Q53 / 155 · WBJEE 2018 · Mathematics
The value of
$I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x,$ is
- A1
- Correct answerB$\pi$
- Ce
- D$\frac{\pi}{2}$
Solution
Let
$I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$
$=\int_{0}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$
$-\int_{0}^{\pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x \ldots(\mathrm{i}$
$I=\int_{0}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\cos x)}}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x$
$\quad-\int_{0}^{\pi / 2} \frac{e^{\tan ^{-1}}(\cos x)}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x \ldots$ (ii)
$\quad\left[\right.$ using, $\left.\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
On adding Eqs. (i) and (ii), we get
$$
\begin{array}{l}
2 I=\int_{0}^{5 \pi / 2} d x-\int_{0}^{\pi / 2} d x \\
\Rightarrow \quad 2 I=(x)_{0}^{5 \pi / 2}-(x)_{0}^{\mathrm{v} / 2} \\
\Rightarrow \quad 2 I=\frac{5 \pi}{2}-\frac{\pi}{2}=2 \pi \quad \Rightarrow \quad I=\pi
\end{array}
$$
Hence, the correct answer is option: 'B'.