I= put tan =x^ 2 ^ 2 =2x dx = 2x x^ 4 +1 dx = 2x^ 2 x^ 4 +1 dx= x^ 2 +1 x^ 4 +1 dx+ x^ 2 -1 x^ 4 +1 dx = 1+1/x^ 2 x^ 2 +1/x^ 2 dx+ 1-1/x^ 2 x^ 2 +1/x^ 2 dx=I_ 1 +I_ 2 I_ 1 = 1+1/x^ 2 x^ 2 +1/x^ 2 dx= 1+1/x^ 2 (x - 1/x)^ 2 +2 dx= dt t^ 2 +2 where x-1 x =t =1 2 ^ -1 t 2 +c_ 1 =1 2 ^ -1 ( x^ 2 -1 2x )+c_ 1 =1 2 ^ -1 ( -1 2 )+c_ 1 I_ 2 = 1-1/x^ 2 x^ 2 +1/x^ 2 dx= 1-1/x^ 2 (x + 1/x)^ 2 -2 dx= dz z^ 2 -2 where x+1 x =z =1 2 ( z-2 z+2 )+c_ 2 =1 22 ( x^ 2 +1- 2x x^ 2 +1- 2x )+c_ 2 =1 22 ( +1-2 +1-2 )+c_ 2 I=1 2 ^ -1 ( -1 2 )+1 22 ( +1-2 +1-2 )2+c.

Evaluate: d .

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Evaluate:

$\int\sqrt{\tan\theta}\text{ d}\theta. $

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Answer

$\text{I}=\text{ }\text{ }\text{ put tan}\theta=\text{x}^{2}\text{ }\text{ }\therefore\text{ }\sec^{2}\theta\text{d}\theta=2\text{x dx}\Rightarrow\text{d}\theta=\frac{\text{2x}}{\text{x}^{4}+1}\text{dx}$
$=\int\frac{\text{2x}^{2}}{\text{x}^{4}+1}\text{dx}=\int\frac{\text{x}^{2}+1}{\text{x}^{4}+1}\text{dx}+\int\frac{\text{x}^{2}-1}{\text{x}^{4}+1}\text{dx }=\int\frac{1+1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}+\int\frac{1-1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\text{I}_{1}+\text{I}_{2} $
$\text{I}_{1}=\int\frac{1+1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\int\frac{1+1/\text{x}^{2}}{\text{(x - 1/x)}^{2}+2}\text{dx}=\int\frac{\text{dt}}{\text{t}^{2}+2}\text{ }\text{ }\text{ }\text{where }\text{x}-\frac{1}{\text{x}}=\text{t}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\text{t}}{\sqrt{2}}+\text{c}_{1}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\text{x}^{2}-1}{\sqrt{2}\text{x}}\Bigg)+\text{c}_{1}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\tan\theta-1}{\sqrt{2\tan\theta}}\Bigg)+\text{c}_{1}$
$\text{I}_{2}=\int\frac{1-1/\text{x}^{2}}{\text{x}^{2}+1/\text{x}^{2}}\text{dx}=\int\frac{1-1/\text{x}^{2}}{\text{(x + 1/x)}^{2}-2}\text{dx}=\frac{\text{dz}}{\text{z}^{2}-2}\text{ }\text{ }\text{ }\text{ where x}+\frac{1}{\text{x}}=\text{z}$
$=\frac{1}{\sqrt{2}}\log\Bigg(\frac{\text{z}-\sqrt{2}}{\text{z}+\sqrt{2}}\Bigg)+\text{c}_{2}=\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\text{x}^{2}+1-\sqrt{\text{2x}}}{\text{x}^{2}+1-\sqrt{\text{2x}}}\Bigg)+\text{c}_{2}=\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\tan\theta+1-\sqrt{2\tan\theta}}{\tan\theta+1-\sqrt{2\tan\theta}}\Bigg)+\text{c}_{2}$
$\therefore\text{ }\text{ }\text{ I}=\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\frac{\tan\theta-1}{\sqrt{2\tan\theta}}\Bigg)+\frac{1}{2\sqrt{2}}\log\Bigg(\frac{\tan\theta+1-\sqrt{2\tan\theta}}{\tan\theta+1-\sqrt{2\tan\theta}}\Bigg)2+\text{c}.$

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