# Inverse Trigonometric Integration

## Reference > Calculus: Integration

 Description$$\int \sin^{-1}{(x)} \, dx=x\sin^{-1}{(x)}+\sqrt{1-{x}^{2}}$$ $$\int \cos^{-1}{(x)} \, dx=x\cos^{-1}{(x)}-\sqrt{1-{x}^{2}}$$ $$\int \tan^{-1}{(x)} \, dx=x\tan^{-1}{(x)}-\frac{1}{2}\ln{(1+{x}^{2})}$$ $$\int \csc^{-1}{(x)} \, dx=x\csc^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}$$ $$\int \sec^{-1}{(x)} \, dx=x\sec^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}$$ $$\int \cot^{-1}{(x)} \, dx=x\cot^{-1}{(x)}+\frac{1}{2}\ln{(1+{x}^{2})}$$
 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$Donearcsin(x)*x+sqrt(1-x^2)+C