Q.Find $\int \dfrac{x \sin^{-1} x}{\sqrt{1 - x^2}}\, dx$

Concept: Power Rule Integration — but the key trick here is a substitution that simplifies the inverse sine term.

Let $t = \sin^{-1} x$. Then $x = \sin t$ and $\frac{dx}{\sqrt{1 - x^2}} = dt$. The integral becomes:

Now integrate by parts: let $u = t$, $dv = \sin t \, dt$, so $du = dt$, $v = -\cos t$.

$\int t \sin t \, dt = -t \cos t + \int \cos t \, dt = -t \cos t + \sin t + C$ …

The key insight here is that the denominator $\sqrt{1 - x^2}$ is a dead giveaway for a trigonometric substitution. When you see $\sqrt{1 - x^2}$, your mind should immediately jump to $x = \sin \theta$ (or $x = \cos \theta$). This substitution will not only simplify the square root but also turn the $\sin^{-1} x$ into something much friendlier — just $\theta$.

Let’s walk through it.

1. Set up the substitution.

   Let $x = \sin \theta$, where $\theta \in [-\pi/2, \pi/2]$ (the principal range of $\sin^{-1}$). Then $dx = \cos \theta \, d\theta$, and $\sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta$ (positive in this range). Also, $\sin^{-1} x = \theta$.

2. Rewrite the integral.

   The numerator becomes $x \sin^{-1} x = (\sin \theta)(\theta)$. The denominator is $\sqrt{1 - x^2} = \cos \theta$. So the integral becomes:

The $\cos \theta$ cancels neatly — that’s the beauty of this substitution.

3. Integrate $\int \theta \sin \theta \, d\theta$. This is a classic product of a polynomial ($\theta$) and a trigonometric function ($\sin \theta$). Use integration by parts. Let $u = \theta$ and $dv = \sin \theta \, d\theta$. Then $du = d\theta$ and $v = -\cos \theta$. Integration by parts gives:

And $\int \cos \theta \, d\theta = \sin \theta$, so:

$\int \theta \sin \theta \, d\theta = -\theta \cos \theta + \sin \theta + C.$ …

Method: Trig Substitution That Simplifies an Inverse Function, Then Parts

Use this when the integrand mixes $\sqrt{1-x^2}$ with $\sin^{-1}x$: the substitution $x=\sin\theta$ cleans up both the root and the inverse trig, after which integration by parts finishes it.

Steps

Step 1: Substitute $x=\sin\theta$.

Then $dx=\cos\theta\,d\theta$, $\sqrt{1-x^2}=\cos\theta$, and crucially $\sin^{-1}x=\theta$. The integrand $\dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx$ collapses to $\int\theta\sin\theta\,d\theta$.

Step 2: Integrate $\theta\sin\theta$ by parts. …

Common Mistakes

Mistake 1: Not recognising $\sqrt{1-x^2}$ as the cue for $x=\sin\theta$.

Why it's wrong: without the substitution the $\sin^{-1}x$ factor stays unmanageable. Correct approach: let $x=\sin\theta$ so $\sin^{-1}x=\theta$.

Mistake 2: Sign slip integrating $\theta\sin\theta$. …