The value of ∫_0^(π)/4(sin x+cos x)/(3+sin 2x) dx is

The correct answer is Option (2) → \(\displaystyle\frac{1}{4}\log_e 3\)

Let \(t = \sin x - \cos x\). Then \(dt = (\cos x + \sin x)\,dx\), which is the numerator.

Squaring t:

    \( t^{2} = 1 - 2\sin x\cos x = 1 - \sin 2x \)
    \( \sin 2x = 1 - t^{2} \)

So the denominator is \(3 + 1 - t^{2} = 4 - t^{2}\).

When \(x = 0\), \(t = -1\); when \(x = \frac{\pi}{4}\), \(t = 0\). So

    \( \displaystyle \int_{0}^{\pi/4}\frac{\sin x+\cos x}{3+\sin 2x}\,dx = \int_{-1}^{0}\frac{dt}{4-t^{2}} \)

Using \(\int\frac{dt}{a^{2}-t^{2}} = \frac{1}{2a}\log_e\left|\frac{a+t}{a-t}\right|\) with \(a = 2\):

    \( \displaystyle = \frac{1}{4}\left[\log_e\left|\frac{2+t}{2-t}\right|\right]_{-1}^{0} \)

    \( \displaystyle = \frac{1}{4}\left(\log_e 1 - \log_e\frac{1}{3}\right) \)

    \( \displaystyle = \frac{1}{4}\log_e 3 \)