Calculus II - Trigonometric Integrals HARD

[GreenPrint]

Homework Statement

Evaluate

integral sin^(-3/2)(x)cos^3(x) dx

Homework Equations

tan(x)=sin(x)/cos(x)
sin^2(x)+cos^2(x)=1
sin^2(x)=(1-cos(x))/2
cos^2(x)=(1+cos(2x))/2
integral cos(x)dx = sin(x) + c
integral sin(x)dx = -cos(x) + c
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)
a^m/a^n=a^(m-n)
a^m*a^n=a^(m+n)
sin(x)=(e^(ix)-e^(-ix))/(2i)
cos(x)=(e^(ix)+e^(-ix))/2
tan(x)=sin(x)/cos(x)
cis(x)=cos(x)+i*sin(x)=e^(ix)
cis(-x)=cos(x)-i*sin(x)=e^(-ix)
e^(i*pi)+1=0

The Attempt at a Solution

I'm at a lost as to how even to being...
I tried using sin^2(x)+cos^2(x)=1
I tried using some of the double angle formulas
every single time I get to the point were I don't know how to proceed

MATLAB answer:
>> int(cos(x)^3/sin(x)^(3/2))
Warning: Explicit integral could not be found.

ans =

int(cos(x)^3/sin(x)^(3/2), x)

Wolfram Alpha answer:
-40/9 F(1/4 (pi-2 x)|2)-2/9 sqrt(sin(x)) (2 cos(x)+3 x (sin(x)+3 csc(x)))+constant

My answer:
Don't even know how to proceed after a couple of steps

Back of the book:
Only odd answers are given... this is problem #16

 

[Ray Vickson]

Homework Statement

Evaluate

integral sin^(-3/2)(x)cos^3(x) dx

Homework Equations

tan(x)=sin(x)/cos(x)
sin^2(x)+cos^2(x)=1
sin^2(x)=(1-cos(x))/2
cos^2(x)=(1+cos(2x))/2
integral cos(x)dx = sin(x) + c
integral sin(x)dx = -cos(x) + c
d/dx sin(x) = cos(x)
d/dx cos(x) = -sin(x)
a^m/a^n=a^(m-n)
a^m*a^n=a^(m+n)
sin(x)=(e^(ix)-e^(-ix))/(2i)
cos(x)=(e^(ix)+e^(-ix))/2
tan(x)=sin(x)/cos(x)
cis(x)=cos(x)+i*sin(x)=e^(ix)
cis(-x)=cos(x)-i*sin(x)=e^(-ix)
e^(i*pi)+1=0

The Attempt at a Solution

I'm at a lost as to how even to being...
I tried using sin^2(x)+cos^2(x)=1
I tried using some of the double angle formulas
every single time I get to the point were I don't know how to proceed

MATLAB answer:
>> int(cos(x)^3/sin(x)^(3/2))
Warning: Explicit integral could not be found.

ans =

int(cos(x)^3/sin(x)^(3/2), x)

Wolfram Alpha answer:
-40/9 F(1/4 (pi-2 x)|2)-2/9 sqrt(sin(x)) (2 cos(x)+3 x (sin(x)+3 csc(x)))+constant

My answer:
Don't even know how to proceed after a couple of steps

Back of the book:
Only odd answers are given... this is problem #16

Maple gets a simple answer:
> f:=cos(x)^3/sin(x)^(3/2);
> J2:=int(f,x);
> J2 := -2/3*(sin(x)^2+3)/sin(x)^(1/2)

We can check to see if dJ2/dx = f:
> simplify(diff(J2,x));

> cos(x)^3/sin(x)^(3/2) <---- correct!

RGV

 

[GreenPrint]

so how do i go about proceeding in this problem?
as far as doing it by hand

 

[Ray Vickson]

Try sin(x)=y.

RGV

 

[I like Serena]

Hi GreenPrint!

I do not know what you did with WolframAlpha, but he gives a simple answer as well.
And with the "Show steps" button, you can see how he got there:
http://www.wolframalpha.com/input/?i=+int%28cos%28x%29^3%2Fsin%28x%29^%283%2F2%29%2C+x%29

 

[eumyang]

What is the original question? The OP wrote it two different ways. Is the integrand this?
\sin^{-3/2} x \text{ }\cos^3 x
Or this?
\frac{\cos^3 x}{\sin^{-3/2} x}

 

[Clever-Name]

Greenprint, there's a simple first step for these sorts of problems.. first:

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

Can you see a substitution you can make now that might make things easier?

edit - eumyang, look at how the fraction version is written. the -3/2 is turned into a 3/2 by putting it in the denominator.