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

Idea: integration by parts with the inverse-trig factor as $u$ (its derivative is algebraic).

Let $u=\cos^{-1}x,\;dv=x\,dx$, so $du=-\dfrac{1}{\sqrt{1-x^2}}\,dx,\;v=\dfrac{x^2}{2}$:

Write $x^2=1-(1-x^2)$: …

Why integration by parts

We have a product of an algebraic factor $x$ and an inverse-trig factor $\cos^{-1}x$. There's no product rule for integrals, so we use integration by parts, $\int u\,dv=uv-\int v\,du$. Choose $u=\cos^{-1}x$ because its derivative $-\tfrac{1}{\sqrt{1-x^2}}$ is algebraic and simplifies the problem.

Step 1 — Apply the formula

Step 2 — The leftover integral

Split the numerator as $x^2=1-(1-x^2)$:

The first is $\sin^{-1}x$; the second is the standard result $\int\sqrt{1-x^2}\,dx=\dfrac{x}{2}\sqrt{1-x^2}+\dfrac12\sin^{-1}x$. Therefore …

Method: Integration by parts with an inverse-cosine factor

For $x^{n}\cos^{-1}x$, differentiate the inverse-cosine (ILATE: Inverse first) to remove it, leaving an algebraic-radical integral.

Steps

Step 1: Assign parts by ILATE.

Step 2: Differentiate and integrate.

Note the minus sign in the derivative of $\cos^{-1}x$ — it is the single most common slip.

Step 3: Apply the formula.

$\int x\cos^{-1}x\,dx=\frac{x^{2}}{2}\cos^{-1}x+\frac12\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx,$ …

Common Mistakes

Mistake 1: Using $\dfrac{d}{dx}\cos^{-1}x=+\dfrac{1}{\sqrt{1-x^{2}}}$.

Why it's wrong: the derivative of $\cos^{-1}x$ carries a negative sign; dropping it flips the sign of every following term. Correct approach: $\dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$.

Mistake 2: Trying to integrate $\dfrac{x^{2}}{\sqrt{1-x^{2}}}$ by naive substitution only. …