Question

Integrate the following function :-
1) $\log x$
2) ${\cos }^{-1}\left(\sqrt{x}\right)$

Answer 1

$\left(i\right)\int \log xdx$

$=\int 1\times \log xdx$

$WeILATE$

$=\log x(\int 1.dx)-\int \frac{1}{x}\int (1-dx)dx$

$\left(\log x\right)\times x-\int \frac{1}{x}\times xdx$

$=x\log x-x+c$

$\left(ii\right)\int {\cos }^{-1}\sqrt{x}dx$

$=\int {\cos }^{-1}\sqrt{x}\times 1\times dx$

I II

$={\cos }^{-1}\sqrt{x}(\int 1.dx)-\int \frac{1}{\sqrt{1-x^2}}(\int 1.dx)dx$

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

let $1-x^2=t^2$

$\therefore -2xdx=2tdt$

$I=x{\cos }^{-1}\sqrt{x}+\int \frac{-tdt}{t}$

$=x{\cos }^{-1}\sqrt{x}-t+c$

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