Answer to Question #108787 in Calculus for Nimra

Question #108787

∫sin⁻¹(ax) dx


1
Expert's answer
2020-04-16T17:23:38-0400

Use integration by parts , let


u=sin1(ax)          dv=dxdu=adx1(ax)2          v=xu=\sin^{-1}(ax) \ \ \ \ \ \ \ \ \ \ dv=dx\\ du=\frac{adx}{\sqrt{1-(ax)^2}} \ \ \ \ \ \ \ \ \ \ v=x

Then

sin1(ax)dx=uvvdu                         =xsin1(ax)axdx1(ax)2                         =xsin1(ax)ax[1(ax)2]1/2dx,use the rule undu=1n+1un+1,     then sin1(ax)dx=xsin1(ax)+1a1(ax)2+C\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


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