The Problem of the Week #331: How do you integrate the arcsine squared function?

[Greg]

Here is this week's POTW:

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Find $\displaystyle\int\arcsin^2(x)\,dx.$

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[Greg]

Congratulations to MarkFL, lfdahl, Opalg and Theia for their correct solutions.

MarkFL's solution:

[sp] Let:

\(\displaystyle I=\int \arcsin^2(x)\,dx\)

Using IBP, I would state:

\(\displaystyle u=\arcsin(x)\implies du=\frac{1}{\sqrt{1-x^2}}\,dx\)

\(\displaystyle dv=\arcsin(x)\,dx\)

To find \(v\), let's use IBP where:

\(\displaystyle r=\arcsin(x)\implies dr=\frac{1}{\sqrt{1-x^2}}\,dx\)

\(\displaystyle ds=dx\implies s=x\)

Hence:

\(\displaystyle v=x\arcsin(x)-\int \frac{x}{\sqrt{1-x^2}}\,dx=x\arcsin(x)+\sqrt{1-x^2}\)

And so we have:

\(\displaystyle I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-\int \left(x\arcsin(x)+\sqrt{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}\,dx\)

\(\displaystyle I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-\int \frac{x}{\sqrt{1-x^2}}\arcsin(x)+1\,dx\)

\(\displaystyle I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-x-\int \frac{x}{\sqrt{1-x^2}}\arcsin(x)\,dx\)

On the remaining integral, use IBP where:

\(\displaystyle u=\arcsin(x)\implies du=\frac{1}{\sqrt{1-x^2}}\,dx\)

\(\displaystyle dv= \frac{x}{\sqrt{1-x^2}}\,dx\implies v=-\sqrt{1-x^2}\)

And so we have:

\(\displaystyle I=\arcsin(x)\left(x\arcsin(x)+\sqrt{1-x^2}\right)-x-\left(-\sqrt{1-x^2}\arcsin(x)+\int \sqrt{1-x^2}\frac{1}{\sqrt{1-x^2}}\,dx\right)\)

And so we conclude:

\(\displaystyle I=\int \arcsin^2(x)\,dx=\arcsin(x)\left(x\arcsin(x)+2\sqrt{1-x^2}\right)-2x+C\)[/sp]

Opalg's solution:

[sp] Let $u = \arcsin(x)$. Then $x = \sin u$, $dx = \cos u\,du$, and \(\displaystyle \int \arcsin^2(x)\,dx = \int u^2\cos u\,du\). Integrate by parts twice to get $$\begin{aligned} \int u^2\cos u\,du &= u^2\sin u - \int2u\sin u\,du \\ &= u^2\sin u + 2u\cos u - \int2\cos u\,du \\ &= u^2\sin u + 2u\cos u - 2\sin u + \text{const.} \end{aligned}$$ Now substitute back in terms of $x$ to get $$\int \arcsin^2(x)\,dx = x(\arcsin^2(x) - 2) + 2\sqrt{1-x^2}\arcsin(x) + \text{const.}$$

(Finally, differentiate that result to check that the derivative is indeed $\arcsin^2(x)$.)[/sp]