∫sin⁻¹(ax) dx
Use integration by parts , let
Then
∫sin−1(ax)dx=uv−∫vdu =xsin−1(ax)−∫axdx1−(ax)2 =xsin−1(ax)−∫ax[1−(ax)2]−1/2dx,use the rule ∫undu=1n+1un+1, then ∫sin−1(ax)dx=xsin−1(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∫sin−1(ax)dx=uv−∫vdu =xsin−1(ax)−∫1−(ax)2axdx =xsin−1(ax)−∫ax[1−(ax)2]−1/2dx,use the rule ∫undu=n+11un+1, then ∫sin−1(ax)dx=xsin−1(ax)+a11−(ax)2+C
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