∫(cos x)/(≤ft(sin x+1)≤ft(sin x+3)) dx is equal to

The correct answer is Option (2) → \(\displaystyle \frac{1}{2}\log\left|\frac{\sin x+1}{\sin x+3}\right|+C\)

Let \(t = \sin x\), so \(dt = \cos x\,dx\) and the integral becomes

    \( \displaystyle \int\frac{dt}{(t+1)(t+3)} \)

By partial fractions:

    \( \displaystyle \frac{1}{(t+1)(t+3)} = \frac{1}{2}\left(\frac{1}{t+1}-\frac{1}{t+3}\right) \)

    \( \displaystyle \int\frac{dt}{(t+1)(t+3)} = \frac{1}{2}\left(\log\left|t+1\right|-\log\left|t+3\right|\right)+C \)

    \( \displaystyle = \frac{1}{2}\log\left|\frac{\sin x+1}{\sin x+3}\right|+C \)