Question

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
(A) 1
(B) \(\pi\)
(C) \(e\)
(D) \(\pi / 2\)

Answer 1

Ans: (B)
$$
\text { Hint : } I=\int_{0}^{\frac{5 x}{2}} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x \quad-\int_{0}^{\frac{1}{2}} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x \Rightarrow I=\frac{5 \pi}{4}-\frac{\pi}{4} \Rightarrow I=\pi
$$

Answer 2

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