WBJEE · Maths · Definite Integration
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 \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\pi\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x\) \(=\int_{0}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x\)…
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