Q.Integrate the following function: $x \sin^{-1}x$

Concept: Integration by parts (LIATE rule — inverse trig comes first).

Let $u = \sin^{-1}x$ and $dv = x\,dx$.

Then $du = \frac{1}{\sqrt{1-x^2}}\,dx$ and $v = \frac{x^2}{2}$.

Using $\int u\,dv = uv - \int v\,du$:

For the remaining integral, substitute $x = \sin\theta$, so $dx = \cos\theta\,d\theta$ and $\sqrt{1-x^2} = \cos\theta$. Then:

$\int \frac{x^2}{\sqrt{1-x^2}}\,dx = \int \sin^2\theta \,d\theta = \int \frac{1-\cos2\theta}{2}\,d\theta = \frac{\theta}{2} - \frac{\sin2\theta}{4} + C$ …

Why this approach works

The integrand is an algebraic function ($x$) times an inverse-trigonometric function ($\sin^{-1}x$). By the LIATE guide we take the inverse-trig factor as the first function, because differentiating $\sin^{-1}x$ removes it, leaving a rational integrand we can handle.

Step-by-step solution

1. Set up integration by parts.

Let $u=\sin^{-1}x$ and $dv=x\,dx$, so

Then

2. Evaluate the remaining integral.

Put $x=\sin\theta$, so $dx=\cos\theta\,d\theta$ and $\sqrt{1-x^2}=\cos\theta$:

Since $\theta=\sin^{-1}x$ and $\sin 2\theta=2x\sqrt{1-x^2}$, …

Method: Integration by parts with an inverse-trig factor

Apply this to $x^{n}\sin^{-1}x$ (an algebraic factor times an inverse trigonometric function), where differentiating the inverse-trig part clears it away.

Steps

Step 1: Choose $u$ = the inverse-trig factor (ILATE puts Inverse first).

Differentiating $\sin^{-1}x$ replaces it with an algebraic expression, turning a hard integrand into a rational one.

Step 2: Form $du$ and $v$.

Step 3: Apply the formula and simplify the leftover. …

Common Mistakes

Mistake 1: Taking $u=x$ and $dv=\sin^{-1}x\,dx$.

Why it's wrong: you cannot integrate $\sin^{-1}x$ to get $v$ without doing by parts first. Correct approach: ILATE — the inverse-trig function is always $u$.

Mistake 2: Writing $\dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{1-x^{2}}$ (missing the square root).

Why it's wrong: the correct derivative is $\dfrac{1}{\sqrt{1-x^{2}}}$; the missing root changes the whole leftover integral. Correct approach: memorise the radical in the denominator. …