Q.Integrate the following function: $(\sin^{-1}x)^2$

Idea: write $(\sin^{-1}x)^2=1\cdot(\sin^{-1}x)^2$ and integrate by parts twice.

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

For the leftover, take $u=\sin^{-1}x,\;dv=\dfrac{x}{\sqrt{1-x^2}}\,dx$, giving $du=\dfrac{dx}{\sqrt{1-x^2}},\;v=-\sqrt{1-x^2}$:

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

The strategy

An inverse-trig function squared has no direct antiderivative. Integration by parts lowers the power: with $dv=dx$ the square becomes our $u$, and each round reduces the exponent of $\sin^{-1}x$ by one until we reach an integral we know.

First integration by parts

Second integration by parts

For $\displaystyle\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx$, choose

(the last using $t=1-x^2$). Then …

Method: Repeated integration by parts to lower a power

Use this when an inverse-trig or log function appears squared (or to a higher power), e.g. $(\sin^{-1}x)^{2}$: by parts with $dv=dx$ knocks the exponent down by one each pass.

Steps

Step 1: Write it as (function)$\times 1$ and take the squared function as $u$.

Step 2: First pass reduces the square to a first power.

Step 3: Second pass on the leftover. …

Common Mistakes

Mistake 1: Differentiating $(\sin^{-1}x)^{2}$ without the chain rule.

Why it's wrong: $\dfrac{d}{dx}(\sin^{-1}x)^{2}=\dfrac{2\sin^{-1}x}{\sqrt{1-x^{2}}}$, not $\dfrac{2\sin^{-1}x}{\dots}$ missing the inner derivative. Correct approach: chain rule gives the extra $\dfrac{1}{\sqrt{1-x^{2}}}$.

Mistake 2: Mishandling the $-2$ when substituting the second result back. …