The value of \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin 2x}{1+\sin^{2}x}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\log_e 2\)
The numerator is the derivative of the denominator, since \(\frac{d}{dx}\left(1+\sin^{2}x\right) = 2\sin x\cos x = \sin 2x\).
Let \(t = 1+\sin^{2}x\). Then \(dt = \sin 2x\,dx\).
When \(x = 0\), \(t = 1\); when \(x = \frac{\pi}{2}\), \(t = 2\). So
\( \displaystyle \int_{0}^{\pi/2}\frac{\sin 2x}{1+\sin^{2}x}\,dx = \int_{1}^{2}\frac{dt}{t} \)
\( \displaystyle = \left[\log_e t\right]_{1}^{2} = \log_e 2 \)
Why the other options are incorrect
Option (2) \(2\log_e 2\) — Incorrect. It takes \(dt = \sin x\cos x\,dx\), leaving an extra factor 2; in fact \(dt = 2\sin x\cos x\,dx = \sin 2x\,dx\).
Option (3) \(\displaystyle\frac{1}{2}\log_e 2\) — Incorrect. It brings in a factor \(\frac{1}{2}\) that is not there, as \(\sin 2x\) is exactly the derivative of \(1+\sin^{2}x\).
Option (4) \(\log_e 3\) — Incorrect. The upper limit of t is \(1 + 1 = 2\), not 3.