Answer to Question #108787 in Calculus for Nimra

Question

∫sin⁻¹(ax) dx

Answer 1

Use integration by parts , let

$$u=\sin^{-1}(ax) \ \ \ \ \ \ \ \ \ \ dv=dx\\ du=\frac{adx}{\sqrt{1-(ax)^2}} \ \ \ \ \ \ \ \ \ \ v=x$$

Then

$\int \sin^{-1}(ax) dx=uv-\int vdu\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x\sin^{-1}(ax)-\int \frac{axdx}{\sqrt{1-(ax)^2}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = x\sin^{-1}(ax)-\int ax[1-(ax)^2 ]^{-1/2}dx, \\ \text{use the rule } \int u^ndu=\frac{1}{n+1}u^{n+1} ,\ \ \ \ \ \text{then }\\ \int \sin^{-1}(ax) dx= x\sin^{-1}(ax)+\frac{1}{a} \sqrt{1-(ax)^2}+C$