Q.Evaluate the definite integral $\int_{0}^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx$

The key idea is to rewrite $\sin 2x$ as $2\sin x \cos x$, then substitute $t = \sin x$ to simplify the integrand.

Let $I = \int_{0}^{\pi/2} \sin 2x \, \tan^{-1}(\sin x) \, dx$.

Since $\sin 2x = 2\sin x \cos x$, we have

Substitute $t = \sin x$, so $dt = \cos x \, dx$. When $x = 0$, $t = 0$; when $x = \pi/2$, $t = 1$.

Thus

Now integrate by parts: let $u = \tan^{-1}(t)$, $dv = 2t \, dt$, so $du = \frac{1}{1+t^2} dt$, $v = t^2$.

Then

At $t=1$, $t^2 \tan^{-1}(1) = 1 \cdot \frac{\pi}{4} = \frac{\pi}{4}$; at $t=0$, it is $0$. …

The first thing to notice is the presence of $\sin 2x$ and $\tan^{-1}(\sin x)$. The function $\sin 2x$ is $2\sin x\cos x$, and $\cos x$ is the derivative of $\sin x$. This is a classic setup for a substitution: let $t = \sin x$. Then $dt = \cos x\,dx$, and the limits change beautifully: when $x = 0$, $t = 0$; when $x = \pi/2$, $t = 1$. The $\sin 2x$ term becomes $2\sin x\cos x\,dx = 2t\,dt$. The integral transforms into something much cleaner.

Let’s work through it step by step.

1. Rewrite the integrand. $\sin 2x = 2\sin x\cos x$, so the integral becomes

2. Substitute $t = \sin x$. Then $dt = \cos x\,dx$, and $\sin x = t$. The factor $2\sin x\cos x\,dx$ becomes $2t\,dt$. The limits: $x=0 \Rightarrow t=0$, $x=\pi/2 \Rightarrow t=1$. So

3. Integrate by parts. We need $\int t \tan^{-1} t\,dt$. Let $u = \tan^{-1} t$ and $dv = t\,dt$. Then $du = \frac{1}{1+t^2}\,dt$ and $v = \frac{t^2}{2}$. Integration by parts gives

4. Simplify the remaining integral.

So

5. Evaluate the definite integral from $0$ to $1$.

Simplify: $2$ times the bracket gives …

Method: Reduce $\sin 2x$ via $t=\sin x$, then integrate by parts

Use this when $\sin 2x$ multiplies a function of $\sin x$: writing $\sin 2x=2\sin x\cos x$ sets up $t=\sin x$, and the remaining $\int t\,\tan^{-1}t\,dt$ is done by parts.

Steps

Step 1: Expand and substitute.

$\sin 2x=2\sin x\cos x$, so with $t=\sin x$ ($dt=\cos x\,dx$) the integral becomes $2\int t\,\tan^{-1}t\,dt$ over the new limits. Do not lose the factor $2$.

Step 2: Integrate by parts. …

Common Mistakes

Mistake 1: Losing the factor $2$ from $\sin 2x$.

Why it's wrong: $\sin 2x=2\sin x\cos x$, so the integral is $2\int_0^1 t\tan^{-1}t\,dt$; dropping the $2$ halves the answer. Correct approach: write $\sin 2x$ out explicitly before substituting.

Mistake 2: Poor by-parts choice.

Why it's wrong: taking $u=t$ instead of $u=\tan^{-1}t$ leaves a harder integral. Correct approach: let $u=\tan^{-1}t$, $dv=t\,dt$. …