Inverse Trigonometric Integration

Reference > Calculus: Integration

Description

∫sin⁡−1(x) dx=xsin⁡−1(x)+1−x2\int \sin^{-1}{(x)} \, dx=x\sin^{-1}{(x)}+\sqrt{1-{x}^{2}}

∫cos⁡−1(x) dx=xcos⁡−1(x)−1−x2\int \cos^{-1}{(x)} \, dx=x\cos^{-1}{(x)}-\sqrt{1-{x}^{2}}

∫tan⁡−1(x) dx=xtan⁡−1(x)−12ln⁡(1+x2)\int \tan^{-1}{(x)} \, dx=x\tan^{-1}{(x)}-\frac{1}{2}\ln{(1+{x}^{2})}

∫csc⁡−1(x) dx=xcsc⁡−1(x)−ln⁡(x+xx−1x2)\int \csc^{-1}{(x)} \, dx=x\csc^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}

∫sec⁡−1(x) dx=xsec⁡−1(x)−ln⁡(x+xx−1x2)\int \sec^{-1}{(x)} \, dx=x\sec^{-1}{(x)}-\ln{(x+x\sqrt{\frac{x-1}{{x}^{2}}})}

∫cot⁡−1(x) dx=xcot⁡−1(x)+12ln⁡(1+x2)\int \cot^{-1}{(x)} \, dx=x\cot^{-1}{(x)}+\frac{1}{2}\ln{(1+{x}^{2})}

Examples
∫sin⁡−1(x) dx\int \sin^{-1}{(x)} \, dx
1
Use Integration by Parts on ∫sin⁡−1(x) dx\int \sin^{-1}{(x)} \, dx.
Let u=sin⁡−1(x)u=\sin^{-1}{(x)}, dv=1dv=1, du=11−x2 dxdu=\frac{1}{\sqrt{1-{x}^{2}}} \, dx, v=xv=x

2
Substitute the above into uv−∫v duuv-\int v \, du.
(sin⁡−1(x))x−∫x1−x2 dx(\sin^{-1}{(x)})x-\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx

3
Use Integration by Substitution on ∫x1−x2 dx\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx.
Let u=1−x2u=1-{x}^{2}, du=−2x dxdu=-2x \, dx, then x dx=−12 dux \, dx=-\frac{1}{2} \, du

4
Using uu and dudu above, rewrite ∫x1−x2 dx\int \frac{x}{\sqrt{1-{x}^{2}}} \, dx.
∫−12u du\int -\frac{1}{2\sqrt{u}} \, du

5
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
−12∫1u du-\frac{1}{2}\int \frac{1}{\sqrt{u}} \, du

6
Since 1x=x−12\frac{1}{\sqrt{x}}={x}^{-\frac{1}{2}}, using the Power Rule, ∫x−12 dx=2x12\int {x}^{-\frac{1}{2}} \, dx=2{x}^{\frac{1}{2}}
−u-\sqrt{u}

7
Substitute u=1−x2u=1-{x}^{2} back into the original integral.
−1−x2-\sqrt{1-{x}^{2}}

8
Rewrite the integral with the completed substitution.
(sin⁡−1(x))x+1−x2(\sin^{-1}{(x)})x+\sqrt{1-{x}^{2}}

9
Add constant.
(sin⁡−1(x))x+1−x2+C(\sin^{-1}{(x)})x+\sqrt{1-{x}^{2}}+C

Done

See Also

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