anonymous one year ago integrate dx/(4+x^2)^2 using trig substituition. im confused i thought it had to be a square root to be used no squared can some explain how toset these up?

1. zepdrix

$\large\rm \int\limits \frac{dx}{(4+x^2)^2}$Ummm trig sub? Yah that'll be fun :)

2. zepdrix

If you factor a 4 out of each term in the denominator you get$\large\rm \int\limits\limits \frac{dx}{(4+x^2)^2}=\frac{1}{4^2}\int\limits\frac{dx}{\left(1+\frac{1}{4}x^2\right)}$Let's bring the 1/4 into the square with the x,$\large\rm =\frac{1}{16}\int\limits\limits\frac{dx}{\left(1+\left[\frac{x}{2}\right]^2\right)}$Now we have that thing in the form: $$\large\rm 1+stuff^2$$ So we can make the substitution: $$\large\rm stuff=\tan\theta$$

3. zepdrix

$\large\rm \frac{x}{2}=\tan\theta$From there, find your $$\large\rm d\theta$$. Plug in all the stuff. Then we'll use some cool trig rules to simplify everything down, and hopefully, be able to integrate easily.

4. zepdrix

From there, find your $$\large\rm dx$$* is what I meant to say. Any trouble finding it? :o

5. zepdrix

Or were my steps confusing?

6. anonymous

Im sorry I just did not understand the find dtheta part. I converted (1+tan0) to sec^20 can i take the derivative of that to dnd d0?

7. zepdrix

$\large\rm =\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\left[\color{orangered}{\frac{x}{2}}\right]^2\right)^{2}}=\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\color{orangered}{\tan^2\theta}\right)^{2}}=\frac{1}{16}\int\limits\frac{dx}{\sec^4\theta}$So you've got that much figured out so far? To find dx, you need to go back to your original substitution. $$\large\rm \frac{x}{2}=\tan\theta$$ Take derivative of that thing.

8. anonymous

so it would turn do x=2tantheta and the derivative is sec^2theta do it would be dtheta over sec^2theta

9. zepdrix

Mmm are you missing a 2 in your dx maybe? Should look like this.$\large\rm x=2\tan theat\qquad\to\qquad \color{royalblue}{dx=2\sec^2\theta~d\theta}$ $\large\rm \frac{1}{16}\int\limits\limits\frac{\color{royalblue}{dx}}{\sec^4\theta}=\frac{1}{16}\int\limits\limits\frac{\color{royalblue}{2\sec^2\theta~d\theta}}{\sec^4\theta}$

10. zepdrix

Ah I made a typo in my x :) lol

11. zepdrix

$\large\rm =\frac{2}{16}\int\limits\frac{d \theta}{\sec^2\theta}$So something like this? Ya you have the right idea.

12. zepdrix

Recall that $\large\rm \cos\theta=\frac{1}{\sec\theta}$

13. anonymous

yeah i have that! and i would just u sub right?!

14. zepdrix

$\large\rm =\frac{1}{8}\int\limits \cos^2\theta~d \theta$ Hmm no. Looks like you'll have to use your Half-Angle Formula at this point. :)

15. anonymous

i got 1/8 * integral of cos^2theta

16. anonymous

$=\frac{ 1 }{ 16 }\int\limits \left( 2\cos ^2 \theta \right)d \theta=\frac{ 1 }{ 16 }\int\limits \left( 1+\cos 2\theta \right) d \theta =?$

17. anonymous

its going to be theta -sin2theta/2 but that would make the denominator for sin 32 and it is wrong @surhithayer

18. zepdrix

No, that sounds fine :) You just need your final answer in terms of $$\large\rm x$$, not $$\large\rm \theta$$. So you need to undo your substitution, perhaps with the use of a triangle.

19. anonymous

$\int\limits \cos 2\theta d \theta =\frac{ \sin 2\theta }{ 2}$

20. anonymous

$=\frac{ 1 }{ 2 }*\frac{ 2 \tan \theta }{ 1-\tan ^2 \theta }=\frac{ \tan \theta }{ 1-\tan ^2 \theta }$