Indefinite Integration (02)

Calculus (Integrals, Series)
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jacks
Posts: 102
Joined: Thu Nov 12, 2015 5:26 pm
Location: Himachal Pradesh (INDIA)

Indefinite Integration (02)

#1

Post by jacks »

Find \[\displaystyle \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}}dx\]
jacks
Posts: 102
Joined: Thu Nov 12, 2015 5:26 pm
Location: Himachal Pradesh (INDIA)

Re: Indefinite Integration (02)

#2

Post by jacks »

I have Tried like this way.. \begin{align*}
\int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}}dx&= \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}}dx\\
&= \int \sqrt{\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\sin ^2 x}+\frac{1}{\sqrt{\cos x}}}}dx\\
&= \int \sqrt{\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{\sqrt{\cos x}+\sin^2 x}{\sin ^2 x.\sqrt{\cos x}}}}dx\\
&=\int \sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x+\sqrt{\cos x}\right).\sqrt{\cos x}}}dx\\
&= \int\sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x .\sqrt{\cos x}+\cos x \right)}}dx\,.
\end{align*}
Let \(\cos x = t^2\) and \(\sin x dx = -2tdt \quad\Leftrightarrow\quad dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}\,.\)
Then
\[\displaystyle = -\int \sqrt{\frac{\left(\sqrt{1-t^4}+t^2 \right).\left(1-t^4\right)}{\left(t.\left(1-t^4\right)\right)+t^2}}.\frac{2t}{\sqrt{1-t^2}}dt\]
Now I am struck here.

How can i proceed after that

Thanks
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