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

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

Using the product-to-sum formula \(2\sin A\cos B = \sin(A+B) + \sin(A-B)\),

    \( \displaystyle \sin 3x\cos x = \frac{1}{2}\left(\sin 4x + \sin 2x\right) \)

So

    \( \displaystyle \int_{0}^{\pi/2}\sin 3x\cos x\,dx = \frac{1}{2}\left[-\frac{\cos 4x}{4} - \frac{\cos 2x}{2}\right]_{0}^{\pi/2} \)

At \(x = \frac{\pi}{2}\), \(\cos 4x = \cos 2\pi = 1\) and \(\cos 2x = \cos\pi = -1\); at \(x = 0\), both are 1:

    \( \displaystyle = \frac{1}{2}\left[\left(-\frac{1}{4} + \frac{1}{2}\right) - \left(-\frac{1}{4} - \frac{1}{2}\right)\right] \)

    \( \displaystyle = \frac{1}{2}\cdot 1 = \frac{1}{2} \)