Question

$\int {\sin }^{-1}2xdx$

Answer 1

$\int {\sin }^{-1}2xdx$

$=\int {\sin }^{-1}2x\times 1dx$

$={\sin }^{-1}2x\int dx-\int \left(\frac{d}{dx}{\sin }^{-1}2x\int dx\right)dx$

$=x{\sin }^{-1}2x-\int \frac{2}{\sqrt{1-4x^2}}\times xdx$

$=x{\sin }^{-1}2x-\int \frac{2x}{\sqrt{1-4x^2}}dx$

$=x{\sin }^{-1}2x+\frac{1}{4}\int {\left(1-4x^2\right)}^{-\frac{1}{2}}\left(-8x\right)dx$

$=x{\sin }^{-1}2x+\frac{1}{4}\times \frac{{\left(1-4x^2\right)}^{\frac{1}{2}}}{{\displaystyle \frac{1}{2}}}+C\left[\begin{array}{c}\int {\left[f\left(x\right)\right]}^{n}f\text{'}\left(x\right)dx=\frac{{\left[f\left(x\right)\right]}^{n+1}}{n+1}+C\\ f\left(x\right)=1-4x^2\Rightarrow f\text{'}\left(x\right)=-8x\end{array}\right]$

$=x{\sin }^{-1}2x+\frac{1}{2}\sqrt{1-4x^2}+C$