MHT CET · Maths · Definite Integration
\(\int_0^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x} d x\) has the value
- A \(-\frac{\pi}{4}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{2}\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int_0^{\pi / 2} \frac{\sin x}{1+\cos ^2 \mathrm{x}} \mathrm{dx}=-\int_1^0 \frac{\mathrm{dt}}{1+\mathrm{t}^2}[\text { let } \cos \mathrm{x}=\mathrm{t}] \\ & =-\left[\tan ^{-1} \mathrm{t}\right]_1^0 \\ & =-\left(0-\frac{\pi}{4}\right) \\ & =\frac{\pi}{4}\end{aligned}\)
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