# Homework Help: Trig integral

1. Apr 10, 2007

### flybyme

1. The problem statement, all variables and given/known data

how would one calculate $$4 \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{(1 + \cos^2 \theta)^2} d \theta$$?

3. The attempt at a solution

someone suggested a $$u = \tan \theta$$ substitution, but i don't understand why and how this would help me. couldn't i just use $$u = \cos t$$?

2. Apr 10, 2007

### AlephZero

Divide top and bottom by $$\cos^4\theta$$.

The function becomes $$\frac{\sec^2\theta}{(\sec^2\theta+1)^2}$$

Does that make $$u=\tan\theta$$ looks like a good move?

u = cos t is probably a bad move, because du = -sin t dt and the sin t doesn't cancel with anything.

Last edited: Apr 10, 2007
3. Apr 10, 2007

### flybyme

the only trig functions we learn here are tan, cos and sin, so i'm not really sure... :)

wouldn't i get du = -sin t dt = -sqrt(1 - cos^2 t) = -sqrt(1 - u^2) dt?

because then the integral should translate to $$-4 \int_1^0 \frac{u^2}{ (u^2+1)(1-u^2)^{1/2} } du$$, or am i wrong?

4. Apr 10, 2007

### drpizza

:surprised

I agree with what you had in the last post, except, wouldn't it be (1+u^2)^2 in the denominator? Anyway, now what?

Last edited: Apr 10, 2007
5. Apr 10, 2007

### flybyme

yeah, i've never understood why that is so...

6. Apr 10, 2007

### mjsd

what is the derivative of $$\tan \theta$$?

7. Apr 10, 2007

### mjsd

also do you know $$1+\tan^2\theta =\sec^2\theta$$?

8. Apr 10, 2007

### drpizza

I edited my post just a little too late... isn't it $$(u^2 + 1)^2$$ in the denominator?

9. Apr 10, 2007

### flybyme

oops. yeah, that's right.

10. Apr 10, 2007

### flybyme

no, i didn't know that. :)

so i would get $$4 \int_0^{\pi/2} \frac{\sec^2 \theta}{(\sec^2 \theta + 1)^2} d \theta = 4 \int_0^{\pi/2} \frac{\tan^2 \theta + 1}{(\tan^2 \theta + 2)^2} d \theta = 4 \int_0^{\Omega} \frac{u^2 + 1}{(u^2 + 2)^2(u^2 + 1)} du = 4 \int_0^{\Omega} \frac{du}{(u^2 + 2)^2}$$

as $$u = \tan \theta$$ and $$du = (1 + \tan^2 \theta) d \theta = (1 + u^2) d \theta$$.

does that seem correct? i'm unsure on what $$\Omega$$ would be though, as $$\tan \pi/2 = \infty$$

Last edited: Apr 10, 2007
11. Apr 11, 2007

### Gib Z

Ahh Nope, $$\int \frac{\sec^2\theta}{(\sec^2\theta+1)^2} d\theta$$ becomes $$\int \frac{1}{u^2} du$$ when u = tan theta, because the derivative of tan theta is (sec theta)^2.

In case you don't know Sec x = 1/(cos x) by definition.

12. Apr 11, 2007

### AlephZero

Oops.... you mean $$\int \frac{1}{(u^2+2)^2} du$$

1 + tan^2 x = sec^2 x

NOT 1 + sec^x = tan^2 x !!!

(And even if that was true, it would have been u^4 not u^2)

13. Apr 12, 2007

### Gib Z

14. Apr 13, 2007

### dextercioby

$$\int_{0}^{\frac{\pi }{2}}\frac{\cos ^{2}x}{\left( 1+\cos ^{2}x\right) ^{2}}dx=\allowbreak \int_{0}^{\infty }\frac{1}{\left( 1+p^{2}\right) ^{2}\left( 1+\frac{1}{1+p^{2}}\right) ^{2}}\,dp=\int_{0}^{\infty }\frac{1}{\left( 2+p^{2}\right) ^{2}}\,dp\allowbreak$$

$$=\allowbreak \int_{0}^{\frac{1}{2}\pi }\frac{\sqrt{2}}{4+4\tan ^{2}q}\,dq=\frac{\sqrt{2}}{4}\int_{0}^{\frac{1}{2}\pi }\cos ^{2}q \ {} \ dq=\frac{\sqrt{2}}{4}\frac{\pi }{4}=\frac{\pi\sqrt{2}}{16}$$

15. Apr 13, 2007

### Gib Z

Would I be the only person to be a tiny bit confused as to what exactly dexter did ...

Last edited: Apr 13, 2007
16. Apr 13, 2007

### dextercioby

Would it help if you knew that $\tan x=p$ and $p=\sqrt{2}\tan q$ ?

17. Apr 13, 2007

### Gib Z

Ahh yup it did :D