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

[chetzread]

Mod note: Based on an attachment in a later post in this thread, the actual expression is
$(9 - 16\cos^2(\theta))^{3/2}$

Homework Statement

: https://www.physicsforums.com/posts/5610105/

Homework Equations

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 ?
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[DrClaude]

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

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?

 

[chetzread]

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?

No, how should it be?

 

[DrClaude]

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

 

[chetzread]

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

Dun know , can you help ?

 

[chetzread]

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

here's my actual problem , i want to 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 ?

 

[Ray Vickson]

here's my actual problem , i want to integrate $$\int_{0.23\pi}^{0.5\pi} \ (9-16cos\theta)^{1.5}\ d\theta$$Is there any other way to integrate this without expanding the terms ?

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.

 

[Mark44]

Thread closed.
@chetzread,
This is what you said:

here's my actual problem , i want to integrate $$\int_{0.23\pi}^{0.5\pi} \ (9-16cos\theta)^{1.5}\ d\theta$$

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.