\(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\cos x}{1+\sin^{2}x}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\pi}{4}\)
Let \(t = \sin x\), so \(dt = \cos x\,dx\).
When \(x = 0\), \(t = 0\); when \(x = \frac{\pi}{2}\), \(t = 1\). Then
\( \displaystyle \int_{0}^{\pi/2}\frac{\cos x}{1+\sin^{2}x}\,dx = \int_{0}^{1}\frac{dt}{1+t^{2}} \)
\( \displaystyle = \left[\tan^{-1}t\right]_{0}^{1} \)
\( \displaystyle = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{2}\) — Incorrect. It uses the x-limit \(\frac{\pi}{2}\) in place of the new upper limit t = 1.
Option (2) \(\log_e 2\) — Incorrect. A log appears for \(\int\frac{2t}{1+t^{2}}\,dt\); here the numerator is just dt, which gives \(\tan^{-1}t\).
Option (4) 1 — Incorrect. The value is \(\frac{\pi}{4} \approx 0.785\), not 1.