Mathematics

Integrate the following
i) $$\log{x}$$
ii) $${\cos}^{-1}(\sqrt{x})$$


SOLUTION
(i)  $$\int_{}^{} {\log xdx} $$
$$ = \int_{}^{} {1.\log xdx} $$
$$ = x\log x - \int_{}^{} {\cfrac{1}{x}.xdx} $$
$$ = x\log x - x + c$$
(ii)  $$\int_{}^{} {{{\cos }^{ - 1}}\sqrt x dx} $$
putting $$x = {t^2}$$
        $$ \Rightarrow dx = 2tdt$$
$$ = 2\int_{}^{} {t{{\cos }^{ - 1}}tdt} $$
$$ = 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \int_{}^{} {\cfrac{{ - 1}}{{\sqrt {1 - {t^2}} }} \times \cfrac{{{t^2}}}{2}dt} } \right]$$
$$ = 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\cfrac{{1 - {t^2} + 1}}{{\sqrt {1 - {t^2}} }}dt} } \right]$$
$$ = 2\left[ {\cfrac{{{t^2}}}{2}{{\cos }^{ - 1}}t - \cfrac{1}{2}\int_{}^{} {\left( {\sqrt {1 - {t^2}}  + \cfrac{1}{{\sqrt {1 - {t^2}} }}} \right)dt} } \right]$$
$$ = {t^2}{\cos ^{ - 1}}t - \int_{}^{} {\sqrt {1 - {t^2}} dt - \int_{}^{} {\cfrac{1}{{\sqrt {1 - {t^2}} }}dt} } $$
$$ = {t^2}{\cos ^{ - 1}}t - \left[ {\cfrac{t}{2}\sqrt {1 - {t^2}}  + \cfrac{1}{2}{{\sin }^{ - 1}}t} \right] - {\sin ^{ - 1}}t + c$$
$$ = {t^2}{\cos ^{ - 1}}t - \cfrac{t}{2}\sqrt {1 - {t^2}}  - \cfrac{3}{2}{\sin ^{ - 1}}t + c$$
$$ = x{\cos ^{ - 1}}\sqrt x  - \cfrac{{\sqrt x }}{2}\sqrt {1 - x}  - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$$
$$ = x{\cos ^{ - 1}}\sqrt x  - \cfrac{{\sqrt {x - {x^2}} }}{2} - \cfrac{3}{2}{\sin {^{ - 1}\sqrt x }} + c$$
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Subjective Medium Published on 17th 09, 2020
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