How Do You Solve the Integral of Cos^5/2(x) Over the Square Root of Sin(x)?

[rocomath]

Integral, hint please

\int\frac{\cos^{\frac{5}{2}}x}{\sqrt{\sin{x}}}dx

Tried: splitting it up, trig identities = no go!

\int\frac{\cos{x}\cos^{\frac{3}{2}}x}{\sqrt{\sin{x}}}dx

Broke up cosine cubed then ended up realizing I would have a cosine left over so that wasn't a good idea.

\int\cos^{2}x\sqrt{\cot{x}}dx

Then my world crumbled.

 

[Rainbow Child]

Try the transformation

x=\arctan(t^2), \quad d\,x=\frac{2\,t}{1+t^4}\,d\,t

 

[coomast]

@ rocophysics: Have you been able to solve this integral? Do you need any information, because it is quite involved and won't post the method if not necessary. It will take me about an hour or so to type everything :-)

 

[rocomath]

@ rocophysics: Have you been able to solve this integral? Do you need any information, because it is quite involved and won't post the method if not necessary. It will take me about an hour or so to type everything :-)

Lol, can I get a hint? I wasn't sure how to apply Rainbow Child's substitution so I decided to put it off for a while. :p

 

[Rainbow Child]

Lol, can I get a hint? I wasn't sure how to apply Rainbow Child's substitution so I decided to put it off for a while. :p

Applying the transformation I wrote, you get

I=\int\frac{2}{(1+t^4)^2}\,d\,t

since

\cos x=\frac{1}{\sqrt{1+\tan^2 x}}\Rightarrow \cos x=\frac{1}{\sqrt{1+t^4}}, \quad \sin x=\sqrt{1-\cos^2x}\Rightarrow \sin x=\frac{t^2}{\sqrt{1+t^4}}

and I think from here is a easy task!

P.S. What about the books, roco?

 

[coomast]

OK, the method that rainbow child proposed was correct. You need to set
t^2=tan(x)
giving
dx=\frac{2t}{1+t^4}dt
Putting this into the integral gives then
I=2 \cdot \int \frac{dt}{(1+t^4)^2}
This one now needs to be solved by expanding the thing into the following partial fraction equation:
\frac{1}{(t^4+1)^2}=\frac{1}{(t^2-\sqrt{2}t+1)^2 \cdot (t^2+\sqrt{2}t+1)^2} = \frac{At+B}{t^2-\sqrt{2}t+1} + \frac{Ct+D}{(t^2-\sqrt{2}t+1)^2} + \frac{Et+F}{t^2+\sqrt{2}t+1} + \frac{Gt+H}{(t^2+\sqrt{2}t+1)^2}
Solving this is now the tedious step :-) You get finally:
A=-\frac{3\sqrt{2}}{16}
B=\frac{3}{8}
C=-\frac{\sqrt{2}}{8}
D=\frac{1}{8}
E=\frac{3\sqrt{2}}{16}
F=\frac{3}{8}
G=\frac{\sqrt{2}}{8}
H=\frac{1}{8}
The remaining integrals are fairly standard but still require a bit of work. Hope this helps, if anything is unclear, let me know.

 

[rocomath]

Amazing, ok let me try that substitution and see how far I can get w/o looking at your soln.

@Rainbow Child, I really appreciate your generosity :-]]] Thanks.