# How do you find the integral of cos^(-1)x dx?

Sep 7, 2016

$= x {\cos}^{- 1} x - \sqrt{1 - {x}^{2}} + C$

#### Explanation:

$\int {\cos}^{- 1} x \mathrm{dx}$

we know $\frac{d}{\mathrm{dx}} \left({\cos}^{- 1} x\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$ so we can try set up an IBP and use that fact

$= \int \frac{d}{\mathrm{dx}} \left(x\right) {\cos}^{- 1} x \mathrm{dx}$

$= x {\cos}^{- 1} x - \int x \frac{d}{\mathrm{dx}} \left({\cos}^{- 1} x\right) \mathrm{dx}$

$= x {\cos}^{- 1} x + \int x \cdot \frac{1}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

we know that $\frac{d}{\mathrm{dx}} \left(\sqrt{1 - {x}^{2}}\right) = \frac{1}{2} \frac{1}{\sqrt{1 - {x}^{2}}} \left(- 2 x\right) = - \frac{x}{\sqrt{1 - {x}^{2}}}$

$= x {\cos}^{- 1} x + \int \frac{d}{\mathrm{dx}} \left(- \sqrt{1 - {x}^{2}}\right) \mathrm{dx}$

$= x {\cos}^{- 1} x - \sqrt{1 - {x}^{2}} + C$