# Integral problem

• May 18th 2010, 12:01 PM
dojo
Integral problem
I'm at a loss with this question: Could I get some pointer please?

By using substitution t = tanx evaluate the following definite integral:

Limits zero to pi/4

I = 1/(9((cosx)^2) - (sinx)^2) dx

Apologies for the notation (or lack of)

I'm stuck at getting this into a solveable integral

D
• May 18th 2010, 12:04 PM
dwsmith
Quote:

Originally Posted by dojo
I'm at a loss with this question: Could I get some pointer please?

By using substitution t = tanx evaluate the following definite integral:

Limits zero to pi/4

I = 1/(9((cosx)^2) - (sinx)^2) dx

Apologies for the notation (or lack of)

I'm stuck at getting this into a solveable integral

D

I guess I means integral.

$\int_{0}^{\frac{\pi}{4}}\left(\frac{1}{9cos^2(x)}-sin^2(x)\right)dx$

Is this what you mean?
• May 18th 2010, 12:12 PM
dwsmith
Well if that is what you meant, I would do this:

$\int_{0}^{\frac{\pi}{4}}\left(\frac{1}{9cos^2(x)}-sin^2(x)\right)dx=\frac{1}{9}\int_{0}^{\frac{\pi}{ 4}}\left(\frac{2}{1+cos(2x)}\right )dx-\int_{0}^{\frac{\pi}{4}}\left(\frac{1-cos(2x)}{2}\right)$

$=\frac{2}{9}\int_{0}^{\frac{\pi}{4}}dx+\frac{1}{9} \int_{0}^{\frac{\pi}{4}}sec(2x)(2dx)-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}dx+...$
• May 18th 2010, 03:34 PM
dojo
No the parenthesis includes all the trig functions as a denominator.
• May 18th 2010, 03:34 PM
tom@ballooncalculus
brackets indicate, rather...

integrate 1&#x2f;&#x28;9 cos&#x5e;2 x - sin&#x5e;2 x&#x29; - Wolfram|Alpha

Click on 'show steps'. Basically, use pythag to write cos in terms of sin, then use the indentity

$\sin x = \frac{2 \tan(\frac{x}{2})}{\tan^2(\frac{x}{2}) + 1}$

then substitute x = 2 arctan u, which is called the Weierstrass sub
• May 18th 2010, 03:38 PM
dojo
Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
• May 18th 2010, 03:46 PM
tom@ballooncalculus
Quote:

Originally Posted by dojo
Yes that is the correct integral. But how do you solve it with the tanx = t substitution???

Yes, they've asked for x = arctan t rather than x = 2 arctan t, so that may be an even better route
• May 18th 2010, 03:52 PM
dojo
Superduper! Thanks (Rock)
• May 18th 2010, 03:52 PM
skeeter
Quote:

Originally Posted by dojo
Yes that is the correct integral. But how do you solve it with the tanx = t substitution???

$t = \tan{x}$

$x = \arctan{t}$

$dx = \frac{dt}{1+t^2}$

$\cos^2{x} = \frac{1}{t^2+1}$

$\sin^2{x} = \frac{t^2}{t^2+1}$

$\int_0^{\frac{\pi}{4}} \frac{dx}{9\cos^2{x} - \sin^2{x}}
$

substitute ...

$\int_0^1 \frac{dt}{9 - t^2}$

use partial fractions to finish