Matematické Fórum

stuart clark · 30. 08. 2013 20:20

$\int\sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}}dx$

stuart clark · 24. 09. 2013 18:05

I have Tried like this way..

$\displaystyle = \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx}$

$\displaystyle = \int \sqrt{\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\sin ^2 x}+\frac{1}{\sqrt{\cos x}}}}dx$

$\displaystyle = \int \sqrt{\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{\sqrt{\cos x}+\sin^2 x}{\sin ^2 x.\sqrt{\cos x}}}}dx$

$\displaystyle =\int \sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x+\sqrt{\cos x}\right).\sqrt{\cos x}}}dx$

$\displaystyle = \int\sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x .\sqrt{\cos x}+\cos x \right)}}dx$

Let $\displaystyle \cos x = t^2$ and $\sin x dx = -2tdt \Leftrightarrow dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}$

$\displaystyle = -\int \sqrt{\frac{\left(\sqrt{1-t^4}+t^2 \right).\left(1-t^4\right)}{\left(t.\left(1-t^4\right)\right)+t^2}}.\frac{2t}{\sqrt{1-t^2}}dt$

Now I am struck here.

So would anyone plz explain me from here

Thanks

Matematické Fórum

#1 30. 08. 2013 20:20

Indefinite Integral (1)

#2 24. 09. 2013 18:05

Re: Indefinite Integral (1)

#3 25. 09. 2013 16:56 Příspěvek uživatele jardofpr byl skryt uživatelem jardofpr. Důvod: mistake

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