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I don't know how to solve this definite integral, maybe the solution is evident but i don't see it :

$\int_0^\frac{\pi}{2} \frac{\cos^3(x)}{(\cos(2x) + \sin(x))}\,dx$

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$$\int \frac{\cos^3 x}{\cos 2x + \sin x}dx = \int \frac{\cos x (1-\sin^2 x)}{1 - 2\sin^2 x + \sin x}dx$$

using trig identities.

Make the substitution $y = \sin x$ and you'll get:

$$\int \frac{1-y^2}{1-2y^2 +y}dy$$

which can be rewritten as:

$$\int \frac{\frac 12 (1-2y^2+y) +\frac 12 (1-y)}{1-2y^2 +y}dy = \frac 12 y + \frac 12 \int\frac{(1-y)}{1-2y^2 +y}dy $$

and a little partial fraction decomposition should quickly resolve it.

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  • $\begingroup$ Thank you, previously I arrived to the passage with 1-y^2 to the numerator and I didn't go on :) Again, thank you :) $\endgroup$ – user132839 Sep 21 '14 at 12:20
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Make the substitution $\sin(x)=t, \cos(x)dx=dt$, so you get:

$$\int_0^\frac{π}{2}\frac{\cos^3x}{\cos(2x)+ \sin(x)}dx=\int_0^\frac{π}{2}\frac{1- \sin^2x}{\cos^2x-\sin^2x+\sin(x)}\cos(x)dx$$

$$\implies\int_0^1\frac{1-t^2}{1-2t^2+t}dt$$

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