1. ## 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

2. 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?

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

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

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

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

5. 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

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

7. 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

8. Superduper! Thanks

9. 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