Integral problem

**dojo** · May 18th 2010, 11:01 AM

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

**dwsmith** · May 18th 2010, 11:04 AM

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?

**dwsmith** · May 18th 2010, 11:12 AM

Well if that is what you meant, I would do this:

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

$=\frac{2}{9}\int_{0}^{\frac{\pi}{4}}dx+\frac{1}{9} \int_{0}^{\frac{\pi}{4}}sec(2x)(2dx)-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}dx+...$

**dojo** · May 18th 2010, 02:34 PM

No the parenthesis includes all the trig functions as a denominator.

**tom@ballooncalculus** · May 18th 2010, 02:34 PM

brackets indicate, rather...

integrate 1/(9 cos^2 x - sin^2 x) - 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

**dojo** · May 18th 2010, 02:38 PM

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

**tom@ballooncalculus** · May 18th 2010, 02:46 PM

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

**dojo** · May 18th 2010, 02:52 PM

Superduper! Thanks

**skeeter** · May 18th 2010, 02:52 PM

Originally Posted by dojo

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

$t = \tan{x}$

$x = \arctan{t}$

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

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

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

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

substitute ...

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

use partial fractions to finish

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