# Change (9-16cos theta) ^1.5 into another form

1. Nov 2, 2016

### chetzread

Mod note: Based on an attachment in a later post in this thread, the actual expression is
$(9 - 16\cos^2(\theta))^{3/2}$
1. The problem statement, all variables and given/known data
: https://www.physicsforums.com/posts/5610105/ [Broken]

2. Relevant equations

3. The attempt at a solution

my working is (9^1.5) - (16^1.5)[ (cos theta)^1.5 ] = 18-64[ (cos theta)^1.5 ] , is it correct ?

Last edited by a moderator: May 8, 2017
2. Nov 2, 2016

### Staff: Mentor

Try setting $\theta = \pi/3$ in ( 9-16cos theta) ^1.5 and (9^1.5) - (16^1.5)[ (cos theta)^1.5 ]. Do you get the same value?

3. Nov 2, 2016

### chetzread

No, how should it be?

4. Nov 2, 2016

### Staff: Mentor

What is $(a+b)^c$?

5. Nov 2, 2016

### chetzread

Dun know , can you help ?

6. Nov 2, 2016

### chetzread

here's my actual problem , i wanna integrate $$\int_{0.23\pi}^{0.5\pi} \ (9-16cos\theta)^{1.5}\ d\theta$$
Mod note: From the attachment, now shown inline, the above should actually be
$\int_{0.23\pi}^{0.5\pi} \ (9-16cos^2\theta)^{1.5}\ d\theta$

Is there any other way to integrate this without expanding the terms ?

Last edited by a moderator: Nov 2, 2016
7. Nov 2, 2016

### Ray Vickson

The answer will be a complex number, because $9 - 16 \cos \theta< 0$ over part of the integration region, so when you take its 3/2-power you get an imaginary number. Besides that, your integration is not "elementary": according to Maple, the indefinite integral of $(a - b \cos\theta)^{3/2}$ involves the so-called Elliptic functions.

Last edited by a moderator: Nov 2, 2016
8. Nov 2, 2016

### Staff: Mentor

Thread closed.
@chetzread,
This is what you said:
The attachment you posted later in this thread shows that the integral is really this: $\int_{0.23\pi}^{0.5\pi} \ (9-16cos^2\theta)^{1.5}\ d\theta$

Because of sloppiness on your part, you wasted a fair amount of time of the people trying to help you. Please start a new thread.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook