Evaluate the following Integral: ∫ sin 2x tan^-1 (sin x) dx, x ∈ [0, π/2]

Question

Evaluate the following Integral: 

\(\int\limits_{0}^{\pi/2}\) sin 2x tan^-1 (sin x) dx

Answer 1

Let I =  \(\int\limits_{0}^{\pi/2}\) sin 2x tan^-1 (sin x) dx

We have sin 2x = 2 sin x cos x

Put sin x = t

⇒ cos x dx = dt (Differentiating both sides)

When x = 0, t = sin 0 = 0

When x = \(\cfrac{\pi}2\), t = sin \(\cfrac{\pi}2\) = 1

So, the new limits are 0 and 1.

Substituting this in the original integral,

We will use integration by parts.

Recall

Here, take f(t) = tan^-1t and g(t) = t

Substituting these values, we evaluate the integral.