Q.Integrate the following function: $\frac{x \cos^{-1}x}{\sqrt{1-x^2}}$

Concept: Power Rule Integration — after a substitution that simplifies the inverse cosine term.

Let $I = \int \frac{x \cos^{-1}x}{\sqrt{1-x^2}} \, dx$.

Step 1: Substitute $t = \cos^{-1}x$. Then $x = \cos t$ and $dx = -\sin t \, dt$. Also $\sqrt{1-x^2} = \sin t$.

Step 2: Rewrite the integral:

Step 3: Integrate by parts: let $u = t$, $dv = \cos t \, dt$, so $du = dt$, $v = \sin t$. Then …

Why this approach works

The factor $\dfrac{x}{\sqrt{1-x^2}}$ integrates cleanly to $-\sqrt{1-x^2}$, and differentiating $\cos^{-1}x$ gives $-\dfrac{1}{\sqrt{1-x^2}}$, which cancels the radical in the leftover integral. So integration by parts collapses the problem completely.

Step-by-step solution

1. Choose the parts.

Let

Then

2. Apply $\int u\,dv = uv-\int v\,du$.

3. Simplify the leftover integral. …

Method: Substitution that turns the integrand into a clean by-parts form

When an inverse-trig function sits over $\sqrt{1-x^{2}}$ with a matching algebraic factor, e.g. $\dfrac{x\cos^{-1}x}{\sqrt{1-x^{2}}}$, substitute the inverse-trig function itself to collapse the radical.

Steps

Step 1: Substitute $t=\cos^{-1}x$ (or integrate by parts directly).

Then $x=\cos t$, $dx=-\sin t\,dt$, and $\sqrt{1-x^{2}}=\sin t$. The awkward radical disappears.

Step 2: Rewrite the integral in $t$.

Step 3: Integrate $t\cos t$ by parts. …

Common Mistakes

Mistake 1: Forgetting the factor $dx=-\sin t\,dt$ under the substitution $t=\cos^{-1}x$.

Why it's wrong: without it the $\sin t$ in the denominator never cancels. Correct approach: the $-\sin t\,dt$ cancels the $\sin t$, leaving $-t\cos t$.

Mistake 2: Missing the leading minus sign.

Why it's wrong: $\dfrac{d}{dx}\cos^{-1}x$ and the $-\sin t\,dt$ both introduce signs; a dropped one flips the answer. Correct approach: track signs to reach $-\sqrt{1-x^{2}}\cos^{-1}x-x+C$. …