Question

$\int {\cos }^{-1}xdx=$

• A. $x{\cos }^{-1}x+\sqrt{1-x^2}+C$
• B. $x{\cos }^{-1}x-\sqrt{1-x^2}+C$
• C. ${\cos }^{-1}x+\sqrt{1-x^2}+C$
• D. None of these

Answer 1

Answer: (B)

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

$\int {\cos }^{-1}xdx$ can be done by using integration by parts

Let $g\left(x\right)=1$ and $f\left(x\right)={\cos }^{-1}x$

By integration by parts , we know that $\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int (f^1\left(x\right)\int g\left(x\right)dx)dx$

$\Rightarrow \int {\cos }^{-1}xdx=x{\cos }^{-1}x-\int x\times (-\frac{1}{\sqrt{1-x^2}})dx=x{\cos }^{-1}x-\sqrt{1-x^2}+C$

Therefore correct option is $B$