Integral problem

Jan 2009
92
1
London
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

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.

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

Is this what you mean?
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Well if that is what you meant, I would do this:

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

\(\displaystyle =\frac{2}{9}\int_{0}^{\frac{\pi}{4}}dx+\frac{1}{9}\int_{0}^{\frac{\pi}{4}}sec(2x)(2dx)-\frac{1}{2}\int_{0}^{\frac{\pi}{4}}dx+...\)
 
Last edited:
Jan 2009
92
1
London
No the parenthesis includes all the trig functions as a denominator.
 
Jan 2009
92
1
London
Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
 
Jan 2009
92
1
London
Superduper! Thanks (Rock)
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Yes that is the correct integral. But how do you solve it with the tanx = t substitution???
\(\displaystyle t = \tan{x}\)

\(\displaystyle x = \arctan{t}\)

\(\displaystyle dx = \frac{dt}{1+t^2} \)

\(\displaystyle \cos^2{x} = \frac{1}{t^2+1}\)

\(\displaystyle \sin^2{x} = \frac{t^2}{t^2+1}\)


\(\displaystyle \int_0^{\frac{\pi}{4}} \frac{dx}{9\cos^2{x} - \sin^2{x}}
\)

substitute ...

\(\displaystyle \int_0^1 \frac{dt}{9 - t^2}\)

use partial fractions to finish