Let I= dx Put =t^2 ^2x dx=2t dt I= 2t ^2x dt=2 t^2 1+t^4 dt = (t^2+1)+(t^2-1) (1+t^4) dt = (t^2+1) 1+t^4 dt+ (t^2-1) 1+t^4 dt = 1+1 t^2 t^2+1 t^2 dt+= 1-1 t^2 t^2+1 t^2 dt = 1- (-1 t^2 )dt (t-1 t )^2+2 + 1+ (-1 t^2 ) (t+1 t )^2-2 dt Put u=t-1 t = (1+1 t^2 )dt and v=t+1 t = (1-1 t^2 )dt I= du u^2+ (2 )^2 + dv v^2- (2 )^2 =1 2 ^ -1 u 2 +1 22 | v-2 v+2 |+C =1 2 ^ -1 ( -1 2 )+1 22 | - 2 +1 + 2 +1 |+C

dx Hint: Put x = t^2

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$\int\sqrt{\tan\text{x}}\text{ dx}$ Hint: Put $\tan x = t^2$

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Answer

Let $\text{I}=\int\sqrt{\tan\text{x}}\text{ dx}$
Put $\tan\text{x}=\text{t}^2$
$\Rightarrow\sec^2\text{x dx}=2\text{t dt}$
$\therefore\ \text{I}=\int\text{t}\cdot\frac{2\text{t}}{\sec^2\text{x}}\text{dt}=2\int\frac{\text{t}^2}{1+\text{t}^4}\text{dt}$
$=\int\frac{(\text{t}^2+1)+(\text{t}^2-1)}{(1+\text{t}^4)}\text{dt}$
$=\int\frac{(\text{t}^2+1)}{1+\text{t}^4}\text{dt}+\frac{(\text{t}^2-1)}{1+\text{t}^4}\text{dt}$
$=\int\frac{1+\frac{1}{\text{t}^2}}{\text{t}^2+\frac{1}{\text{t}^2}}\text{dt}+=\int\frac{1-\frac{1}{\text{t}^2}}{\text{t}^2+\frac{1}{\text{t}^2}}\text{dt}$
$=\int\frac{1-\Big(-\frac{1}{\text{t}^2}\Big)\text{dt}}{\Big(\text{t}-\frac{1}{\text{t}}\Big)^2+2}+\int\frac{1+\Big(-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}+\frac{1}{\text{t}}\Big)^2-2}\text{dt}$
Put $\text{u}=\text{t}-\frac{1}{\text{t}}$
$\Rightarrow\text{du}=\Big(1+\frac{1}{\text{t}^2}\Big)\text{dt}$
and $\text{v}=\text{t}+\frac{1}{\text{t}}$
$\Rightarrow\text{dv}=\Big(1-\frac{1}{\text{t}^2}\Big)\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{du}}{\text{u}^2+\big(\sqrt{2}\big)^2}+\int\frac{\text{dv}}{\text{v}^2-\big(\sqrt{2}\big)^2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\text{u}}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{v}-\sqrt{2}}{\text{v}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\tan\text{x}-1}{\sqrt{2\tan\text{x}}}\Big)+\frac{1}{2\sqrt{2}}\log\bigg|\frac{\tan\text{x}-\sqrt{2\tan\text{x}}+1}{\tan\text{x}+\sqrt{2\tan\text{x}}+1}\bigg|+\text{C}$

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