Question

Evaluate the following integrals :
$\displaystyle \sf \int \sqrt{x} \: tan \bigg \{2 {tan}^{ - 1} \bigg( \dfrac{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: - \sqrt{ \sqrt{1 + \sqrt{x} } - 1 } }{ \sqrt{ \sqrt{1 + \sqrt{x}} + 1} \: + \sqrt{ \sqrt{1 + \sqrt{x} } - 1 }} \bigg) \bigg \} \: dx$
​

Answer 1

Consider the fraction,

$\longrightarrow F=\dfrac{\sqrt{\sqrt{1+\sqrt x}+1}-\sqrt{\sqrt{1+\sqrt x}-1}}{\sqrt{\sqrt{1+\sqrt x}+1}+\sqrt{\sqrt{1+\sqrt x}-1}}$

Dividing both numerator and denominator by $\sqrt{\sqrt{1+\sqrt x}+1},$

$\longrightarrow F=\dfrac{\left(1-\sqrt{\dfrac{\sqrt{1+\sqrt x}-1}{\sqrt{1+\sqrt x}+1}}\right)}{\left(1+\sqrt{\dfrac{\sqrt{1+\sqrt x}-1}{\sqrt{1+\sqrt x}+1}}\right)}\quad\quad\dots(1)$

Put,

$\longrightarrow\sqrt{\dfrac{\sqrt{1+\sqrt x}-1}{\sqrt{1+\sqrt x}+1}}=\tan\theta$

$\longrightarrow\dfrac{\sqrt{1+\sqrt x}-1}{\sqrt{1+\sqrt x}+1}=\tan^2\theta$

By rule of componendo and dividendo,

$\longrightarrow\sqrt{1+\sqrt x}=\dfrac{1+\tan^2\theta}{1-\tan^2\theta}$

$\longrightarrow\sqrt{1+\sqrt x}=\sec(2\theta)\quad\left[\because\cos(2\theta)=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right]$

$\longrightarrow1+\sqrt x=\sec^2(2\theta)$

$\longrightarrow\sqrt x=\sec^2(2\theta)-1$

$\longrightarrow\sqrt x=\tan^2(2\theta)\quad[\because\sec^2(2\theta)-\tan^2(2\theta)=1]$

$\longrightarrow x=\tan^4(2\theta)$

$\longrightarrow dx=4\tan^3(2\theta)\cdot\sec^2(2\theta)\cdot2\ d\theta$

$\longrightarrow dx=8\tan^3(2\theta)\sec^2(2\theta)\ d\theta$

Then (1) becomes,

$\longrightarrow F=\dfrac{1-\tan\theta}{1+\tan\theta}$

$\longrightarrow F=\tan\left(\dfrac{\pi}{4}-\theta\right)$

We're given the integral,

$\displaystyle\longrightarrow I=\int\sqrt x\tan\left[2\tan^{-1}\left(\dfrac{\sqrt{\sqrt{1+\sqrt x}+1}-\sqrt{\sqrt{1+\sqrt x}-1}}{\sqrt{\sqrt{1+\sqrt x}+1}+\sqrt{\sqrt{1+\sqrt x}-1}}\right)\right]\ dx$

which becomes,

$\displaystyle\longrightarrow I=\int\tan^2(2\theta)\tan\left[2\tan^{-1}\left(\tan\left(\dfrac{\pi}{4}-\theta\right)\right)\right]\cdot 8\tan^3(2\theta)\sec^2(2\theta)\ d\theta$

$\displaystyle\longrightarrow I=8\int\tan^5(2\theta)\tan\left[2\left(\dfrac{\pi}{4}-\theta\right)\right]\sec^2(2\theta)\ d\theta$

$\displaystyle\longrightarrow I=8\int\tan^5(2\theta)\tan\left(\dfrac{\pi}{2}-2\theta\right)\sec^2(2\theta)\ d\theta$

$\displaystyle\longrightarrow I=8\int\tan^5(2\theta)\cot(2\theta)\sec^2(2\theta)\ d\theta$

$\displaystyle\longrightarrow I=4\int\tan^4(2\theta)\cdot 2\sec^2(2\theta)\ d\theta\quad\quad\dots(2)$

Put,

$\longrightarrow u=\tan(2\theta)$

$\longrightarrow du=2\sec^2(2\theta)\ d\theta$

Then (2) becomes,

$\displaystyle\longrightarrow I=4\int u^4\ du$

$\displaystyle\longrightarrow I=\dfrac{4}{5}\,u^5+C$

Undoing $u=\tan(2\theta),$

$\displaystyle\longrightarrow I=\dfrac{4}{5}\tan^5(2\theta)+C$

$\displaystyle\longrightarrow I=\dfrac{4}{5}\left(\tan^4(2\theta)\right)^{\frac{5}{4}}+C$

Undoing $x=\tan^4(2\theta),$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{4}{5}\,x^{\frac{5}{4}}+C}}$

Answer 2

plz mark me as brainlist