|
English
|
Inverse Trigonometric IntegrationDescription \(\int \sin^{-1}{(x)} \, dx=x\sin^{-1}{(x)}+\sqrt{1-{x}^{2}}\) \(\int \cos^{-1}{(x)} \, dx=x\cos^{-1}{(x)}-\sqrt{1-{x}^{2}}\) \(\int \tan^{-1}{(x)} \, dx=x\tan^{-1}{(x)}-\frac{1}{2}\ln{(1+{x}^{2})}\) \(\int \csc^{-1}{(x)} \, dx=x\csc^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}\) \(\int \sec^{-1}{(x)} \, dx=x\sec^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}\) \(\int \cot^{-1}{(x)} \, dx=x\cot^{-1}{(x)}+\frac{1}{2}\ln{(1+{x}^{2})}\) |
Examples \[\int \sin^{-1}{(x)} \, dx\] 1 Use Integration by Parts  on \(\int \sin^{-1}{(x)} \, dx\)Let \(u=\sin^{-1}{(x)}\), \(dv=1\), \(du=\frac{1}{\sqrt{1-{x}^{2}}} \, dx\), \(v=x\) 2 Substitute the above into \(uv-\int v \, du\) \[(\sin^{-1}{(x)})x-\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\] 3 Use Integration by Substitution on \(\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\)Let \(u=1-{x}^{2}\), \(du=-2x \, dx\), then \(x \, dx=-\frac{1}{2} \, du\) 4 Using \(u\) and \(du\) above, rewrite \(\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx\) \[\int -\frac{1}{2\sqrt{u}} \, du\] 5 Use Constant Factor Rule : \(\int cf(x) \, dx=c\int f(x) \, dx\)\[-\frac{1}{2}\int \frac{1}{\sqrt{u}} \, du\] 6 Since \(\frac{1}{\sqrt{x}}={x}^{-\frac{1}{2}}\), using the Power Rule , \(\int {x}^{-\frac{1}{2}} \, dx=2{x}^{\frac{1}{2}}\)\[-\sqrt{u}\] 7 Substitute \(u=1-{x}^{2}\) back into the original integral \[-\sqrt{1-{x}^{2}}\] 8 Rewrite the integral with the completed substitution \[(\sin^{-1}{(x)})x+\sqrt{1-{x}^{2}}\] 9 Add constant \[(\sin^{-1}{(x)})x+\sqrt{1-{x}^{2}}+C\] Done  |
See Also |