∫(sin 2x)/(4sin^2x+9cos^2x) dx is equal to

The correct answer is Option (1) → \(\displaystyle -\frac{1}{5}\log\left(4\sin^2x+9\cos^2x\right)+C\)

Let \(t = 4\sin^2x+9\cos^2x\). Its derivative is a multiple of the numerator:

    \( \displaystyle \frac{dt}{dx} = 8\sin x\cos x-18\sin x\cos x \)

    \( = -10\sin x\cos x = -5\sin 2x \)

So \(\sin 2x\,dx = -\frac{1}{5}\,dt\), and

    \( \displaystyle \int\frac{\sin 2x}{4\sin^2x+9\cos^2x}\,dx = -\frac{1}{5}\int\frac{dt}{t} \)

    \( \displaystyle = -\frac{1}{5}\log\left(4\sin^2x+9\cos^2x\right)+C \)

No modulus is needed because \(4\sin^2x+9\cos^2x \gt 0\) for every x.