Question

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?

Answer 1

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

Answer 2

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: \(\large\rm stuff=\tan\theta\)

Answer 3

\[\large\rm \frac{x}{2}=\tan\theta\]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.

Answer 4

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

Answer 5

Or were my steps confusing?

Answer 6

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?

Answer 7

\[\large\rm =\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\left[\color{orangered}{\frac{x}{2}}\right]^2\right)^{2}}=\frac{1}{16}\int\limits\limits\limits\limits\frac{dx}{\left(1+\color{orangered}{\tan^2\theta}\right)^{2}}=\frac{1}{16}\int\limits\frac{dx}{\sec^4\theta}\]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.

Answer 8

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

Answer 9

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

Answer 10

Ah I made a typo in my x :) lol

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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.

Answer 19

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

Answer 20

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