Find the integral $\int\frac{\arcsin x}{(1-x^2)^\frac32} dx$

Question

Find the integral $$\int\dfrac{\arcsin x}{(1-x^2)^\frac32} dx$$

We can see that $d(\arcsin x)=\dfrac{1}{\sqrt{1-x^2}} dx$.

So we can write the given integral as $$\int\dfrac{\arcsin x}{1-x^2}d(\arcsin x),$$ which I didn't find very helpful.

Another thing I tried is to put $x=\sin t \Rightarrow t=\arcsin x$. Then the given integral can be written as $$\int\dfrac{t}{\cos^3 t}d\sin t,$$ which I also don't know how to find.

Any help will be appreciated. Thanks!

Answer 1

Note that $\mathrm d\sin t=\cos t\,\mathrm dt$; so,\begin{align}\int\frac t{\cos^3t}\mathrm d\sin t&=\int\frac t{\cos^2t}\,\mathrm dt\\&=t\tan t-\int\tan t\,\mathrm dt\\&=t\tan t+\log(\cos t).\end{align}And therefore$$\int\frac{\arcsin x}{(1-x^2)^{3/2}}\,\mathrm dx=\frac{x\arcsin(x)}{\sqrt{1-x^2}}+\log\left(\sqrt{1-x^2}\right).$$

Answer 2

Let $t = \arcsin(x)$. Then $x = \sin(t)$, and \begin{align} \int \frac{\arcsin x}{(1-x^2)^\frac32} \, dx &= \int \frac{t}{(1-x^2)^\frac32} \sqrt{1-x^2} \, dt \\ &= \int \frac{t}{(1-x^2)} \, dt \\ &= \int \frac{t}{1-\sin^2 t} \, dt \\ &= \int \frac{t}{\cos^2 t} \, dt \\ &= \int t\sec^2 t \, dt. \end{align} Then, using integration by parts, $u = t \implies du = dt$ and $dv = \sec^2 t \, dt \implies v = \tan t$, and \begin{align} \int t\sec^2 t \, dt &= t\tan t - \int \tan t dt \\ &= t\tan t - (-\ln(\cos t)) + c \\ &= t\tan t + \ln(\cos t) + c. \end{align} And substituting $t = \arcsin(x)$ back in, we get, \begin{align} \int \frac{\arcsin x}{(1-x^2)^\frac32} \, dx &= \arcsin x \cdot \tan (\arcsin x) + \ln(\cos (\arcsin x )) + c \\ &= \arcsin x \cdot \frac{x}{\cos(\arcsin x)} + \ln(\cos (\arcsin x )) + c. \end{align}