Trigonometric integral problem

sahen
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Homework Statement


\int\frac{1+\sqrt{cosx}}{sinx} Hi , I really need a help for this


Homework Equations





The Attempt at a Solution



\int\frac{1+\sqrt{cosx}}{\sqrt{sin}\sqrt{sin}}

\int\frac{1}{sinx} + \int\frac{1}{\sqrt{sinx}}\sqrt{}\frac{cosx}{sinx} =

\int cscx + \int\sqrt{cscx cotx} = ?

\int cscx is alright but i have no idea for second part .

do you have any idea ?
 
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Make the substitution \sqrt{cosx}=a Then simplify the resulting expression, use partial fractions and you're done.
 
Yes that's it! thanks for your help...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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