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

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The discussion focuses on finding the integral of the arcsine squared function, specifically ∫arcsin²(x)dx. Two main solutions are presented: MarkFL's approach uses integration by parts (IBP) and leads to a complex expression involving arcsin and square root terms, ultimately simplifying to arcsin(x)(xarcsin(x) + 2√(1-x²)) - 2x + C. Opalg's method also employs IBP, transforming the integral into a function of u, resulting in a final expression that includes terms of x and arcsin(x). Both solutions confirm the correctness of their results through differentiation.
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Here is this week's POTW:

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

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to MarkFL, lfdahl, Opalg and Theia for their correct solutions.

MarkFL's solution:

[sp] Let:

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

Using IBP, I would state:

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

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

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

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

$$ds=dx\implies s=x$$

Hence:

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

And so we have:

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

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

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

On the remaining integral, use IBP where:

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

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

And so we have:

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

And so we conclude:

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

Opalg's solution:

[sp] Let $u = \arcsin(x)$. Then $x = \sin u$, $dx = \cos u\,du$, and $$\int \arcsin^2(x)\,dx = \int u^2\cos u\,du$$. 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]
 
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