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I would like to know if there is some way to calculate exactly

$$\int_{0}^{\pi/2} \frac{\cos^3 x}{\sqrt {\sin x}(1+\sin^3 x)}\,dx$$

The numerical value is 1,52069

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Mathematica gives the exact answer as $\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}$. – Zev Chonoles Dec 25 '11 at 19:05
Try substituting $y = \sqrt{\sin(x)}$ which changes the integral to $$\int_0^1\frac{1-y^4}{1+y^6}\mathrm dx.$$ Does that help any? – Dilip Sarwate Dec 25 '11 at 19:07
Since $\cos(x)$ only appears at an odd power, the standard substitution $u=\sin(x)$ works... – N. S. Dec 25 '11 at 19:38
@DilipSarwate I think you need to change the $dx$ to $dy$ and the integral in question is twice your integral. – user21436 Dec 25 '11 at 19:41
@KannappanSampath Yes, you are correct. Unfortunately, comments cannot be edited except right away... – Dilip Sarwate Dec 25 '11 at 20:00

From Dilip Sarwate's suggestion on:

$$2\int_0^1\frac{1-y^4}{1+y^6}\mathrm dy = 2\int_0^1\frac{(1-y^2)(1+y^2)}{(1+y^2)(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy \; .$$

The denominator can be split further into

$$2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^2+\sqrt{3}y+1)(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$

Using partial fractions this becomes

$$2\int_0^1\frac{(1-y^2)}{(y^2-\sqrt{3}y+1)(y^2+\sqrt{3}y+1)}\mathrm dy = 2\int_0^1\frac{\frac{y}{\sqrt{3}}+\frac{1}{2}}{(y^2+\sqrt{3}y+1)}-\frac{\frac{y}{\sqrt{3}}-\frac{1}{2}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$

The last part can be rewritten as

$$\frac{1}{\sqrt{3}}\int_0^1\frac{2y+\sqrt{3}}{(y^2+\sqrt{3}y+1)}-\frac{2y-\sqrt{3}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$

And can easily be solved by substitution to give as Zev Chonoles already mentioned in the commentary: $\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}$ .

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I am sure this substitution might work. Let $u=\sin x$ so that $du=\cos xdx$. Then your integrand becomes; $$\int_0^{\pi/2}\frac{\cos^2x\cos xdx}{\sqrt \sin x(1+\sin^3x)}=\int_0^{\pi/2}\frac{(1-\sin^2x)\cos xdx}{\sqrt \sin x(1+\sin^3x)}=\int_0^1\frac{(1-u^2)du}{\sqrt u(1+u^3)}$$ $$=\int_0^1\frac{(1-u)(1+u)du}{\sqrt u(1+u)(1-u+u^2)}=\int_0^1\frac{(1-u)du}{\sqrt u(1-u+u^2)}$$ I believe at this point you can employ the method of partial fractions to integrate. HINT: Let $u=v^2$ so that $du=2vdv$. Then you have $$\int_0^1\frac{(1-u)du}{\sqrt u(1-u+u^2)}=\int_0^1\frac{2(1-v^2)dv}{(1-v^2+v^4)}$$ and partial fraction follows. I think at this point the procedure for partial fraction is the same as above.

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Thank you Dilip. The substitution works indeed! With partial fraction (not at all nice) the exact value is $\frac{2}{\sqrt3} ln(2+\sqrt3)$ which is exactly the result Zev gave us. – Martin Dec 25 '11 at 19:59

If I were in front of a classroom and a student asked me about this one, I'd begin by writing this: $$\int_{0}^{\pi/2} \frac{\cos^3 x}{\sqrt {\sin x}(1+\sin^3 x)}\,dx = \int_{0}^{\pi/2} \frac{\cos^2 x}{\sqrt {\sin x}(1+\sin^3 x)}{\Huge(}\cos x \;dx{\Huge)}$$ If they don't know how to construe that as a hint, I'd tell them that's what they need to work on.

The point is that $$ du = \cos x\; dx, $$ so that suggests $u=\sin x$.

On the bottom we've got $\sqrt{u}(1+u^3)$. On top we need to change $\cos^2 x$ to $1-\sin^2 x$ so that we'll have $1-u^2$.

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Very well explained Michael. Your reasoning really motivates the substitution and makes it natural. – Martin Dec 25 '11 at 22:08
"On the bottom we've got $\sqrt{u}(1+u^3)$" which is what smanoos ended up with, and then put $u = v^2$ to get to the form that Raskolnikov worked with. So a better suggestion might have been to ask the students to think of how that pesky $\sqrt{\sin x}$ might be eliminated and to suggest looking at $\frac{\mathrm d}{\mathrm dx}\sqrt{\sin x}$ right away. – Dilip Sarwate Dec 25 '11 at 22:14

After we deduce our integral is equal to $$I= 2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy$$ we can finish off like this: $$ I = -2 \int^1_0 \frac{ y^2 (1 - 1/y^2) }{ y^2 (y^2-1 + 1/y^2) } dy = -2 \lim_{a\to 0^+} \int^1_a \frac{ d(y+1/y) }{(y+1/y)^2-3} =2 \int_2^{\infty} \frac{du}{u^2-3}=\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}.$$

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