JEE Advanced Indefinite Integration 1

$I=\displaystyle \int \frac { \sqrt { 1-\sqrt { x } } }{ \sqrt { 1+\sqrt { x } } } dx\\ x=\cos ^{ 2 }{ t } \\ dx=-\sin { 2t } dt\\ \frac { \sqrt { 1-\sqrt { x } } }{ \sqrt { 1+\sqrt { x } } } =\frac { \sqrt { 1-\cos { t } } }{ \sqrt { 1+\cos { t } } } =\tan { \frac { t }{ 2 } } =\frac { \sin { t } }{ 1+\cos { t } } \\ Substituting\quad the\quad last\quad value\\ I=\displaystyle -\int \frac { \sin { t } }{ 1+\cos { t } } \sin { 2t } dt\\ I=\displaystyle -\int \frac { 2\sin ^{ 2 }{ t } \cos { t } }{ 1+\cos { t } } dt\\ I=\displaystyle -\int \frac { 2(1+\cos { t } )(1-\cos { t } )\cos { t } }{ 1+\cos { t } } dt\\ I=\displaystyle \int 2(\cos { t } -1)\cos { t } dt\\ I=\displaystyle \int (1+\cos { 2t } )-2\cos { t } dt\\ I=\displaystyle t+\frac { \sin { 2t } }{ 2 } -2\sin { t } +C\\ Substituting\quad \cos { x } =\sqrt { x } \\ I=\displaystyle \boxed { \boxed { (\sqrt { x } -2)\sqrt { 1-x } +\cos ^{ -1 }{ x } +C } }$

Looking at the integral, first step has to remove nested radical. So I substituted $x=t^2$ ( $dx=2tdt$ ) first. It becomes $\int 2t\sqrt{\dfrac{1-t}{1+t}} dt$

Now, rationalize the numerator by multiplying by $\sqrt{1-t}$ which results in $2\int\dfrac{t(1-t)}{\sqrt{1-t^2}} dt$ $2\int\dfrac{t-1+(1-t^2)}{\sqrt{1-t^2}} dt$ $=2\int\dfrac{t}{\sqrt{1-t^2}} dt-2\int\dfrac{dt}{\sqrt{1-t^2}}+2\int\sqrt{1-t^2} dt$

$=-2\sqrt{1-t^2}-2\sin^{-1}t+2\left [\dfrac{t}{2}\sqrt{1-t^2}+\dfrac{1}{2}\sin^{-1} t\right ] +C$ $=(t-2)\sqrt{1-t^2}-\sin^{-1}y+C$ $=(\sqrt{x}-2)\sqrt{1-x}+\cos^{-1}x+C'$

ok let us begin by multiplying root(1-root(x)) up and down, and now follow me

$\sqrt { \frac { 1-\sqrt { x } }{ 1+\sqrt { x } } } =\quad \frac { 1-\sqrt { x } }{ \sqrt { 1-x } } =\quad \frac { 1 }{ \sqrt { 1-x } } -\sqrt { \frac { x }{ 1-x } }$

Now the first part is easy, let us focus on second, let us substitute

$x={ (sin\theta ) }^{ 2 }\\ \\ dx=\quad 2sin\theta \quad cos\theta \quad d\theta \\ \\ \sqrt { \frac { x }{ 1-x } } =\frac { 2sin\quad sin\quad cos }{ cos } d\theta \quad =\quad 2{ sin }^{ 2 }\quad d\theta \quad =\quad 1-\quad cos2\theta \quad \\ \\$

which can be easily integrated to get

$\\ \theta -\frac { 1 }{ 2 } sin(2\theta )\quad +c\quad \\ \\ =arcsin(\sqrt { x } )\quad -\quad \sqrt { x } \sqrt { 1-x\quad } +c\quad =\quad -arccos(\sqrt { x } )\quad -\quad \sqrt { x } \sqrt { 1-x\quad } \quad +c\quad +\pi /2\quad \\ \\ -arccos(\sqrt { x } )\quad -\quad \sqrt { x } \sqrt { 1-x\quad } \quad +\quad d\\ \\ \\$

the whole integral is (combining first easy part and second needed substitution part becomes

$-2\sqrt { 1-x }+arccos(\sqrt { x } )\quad +\quad \sqrt { x } \sqrt { 1-x } \quad =\quad \left( -2+\sqrt { x } \right) \sqrt { 1-x } +arccos(\sqrt { x } )$

i hope it is clear