The value of ∫_0^(π)/2(cos x)/((1 + sin x)(2 + sin x)) dx is

The correct answer is Option (1) → \(\log_e\displaystyle\frac{4}{3}\)

Let \(t = \sin x\). Then \(dt = \cos x\,dx\), and the limits become

    \( x = 0 \Rightarrow t = 0 \)
    \( \displaystyle x = \frac{\pi}{2} \Rightarrow t = 1 \)

So

    \( \displaystyle I = \int_{0}^{1}\frac{dt}{(1 + t)(2 + t)} \)

By partial fractions:

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

Integrating:

    \( \displaystyle I = \Big[\log_e(1 + t) - \log_e(2 + t)\Big]_{0}^{1} \)

    \( = (\log_e 2 - \log_e 3) - (0 - \log_e 2) \)

    \( = 2\log_e 2 - \log_e 3 \)

    \( \displaystyle = \log_e\frac{4}{3} \)