dx Hint: Put tanx = t^2 - Vidyadip

Question

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

✓

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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