Help to find integral from pi/2 to 2arctg2 of dx/((sin^2x)(1-cosx))

[mobimore]

Integrate from pi/2 to 2arctg2 dx/((sin^2x)(1-cosx))
I tried to substitute sin^2x=t and got integral of dt/(t*sqrt(1-t^2))
Is it right? How can i solve it.

 

[ksdhart2]

Well, first off, the problem is not 100% clear to me. I believe this is what you were assigned. Please correct me if I am mistaken:

\(\displaystyle \displaystyle \int _{\frac{\pi}{2}}^{2tan^{-1}\left(2\right)}\: \frac{1}{sin^2\left(x\right)\cdot\left(1-cos\left(x\right)\right)}dx\)

If the above is the correct problem, I'd try integration by parts. You have:

\(\displaystyle \displaystyle \int _{\frac{\pi }{2}}^{2tan^{-1}\left(2\right)}\:\frac{1}{sin^2\left(x\right)} \cdot \frac{1}{1-cos\left(x\right)}dx\)

\(\displaystyle \displaystyle = \int _{\frac{\pi }{2}}^{2tan^{-1}\left(2\right)}\: \frac{1}{1-cos\left(x\right)} \cdot csc^2\left(x\right)dx\)

Let \(\displaystyle \displaystyle u=\frac{1}{1-cos\left(x\right)}\) and \(\displaystyle dv=csc^2\left(x\right)dx\) and try to continue from here...

 

[Ishuda]

Integrate from pi/2 to 2arctg2 dx/((sin^2x)(1-cosx))
I tried to substitute sin^2x=t and got integral of dt/(t*sqrt(1-t^2))
Is it right? How can i solve it.

I don't get that.
If
t = sin²(x)
then
dt = 2 sin(x) cos(x) dx
or
dx = \(\displaystyle \dfrac{dt}{2\, sin(x)\, cos(x)}\)
sin(x) = t^1/2
cos(x) = (1 - t)^1/2

EDIT: BTW, I probably would have used t = sin(x)

 

[mobimore]

Thanks. I think I solved it!