Question

evaluate the integral 16(1/((x^2+4)^2)) from 0 to 2

Answer 1

\[16\int\limits_{0}^{2}\frac{ 1 }{ (x^2+4)^2 }\]

Answer 2

I thought trig substitution was probably a good route?

Answer 3

the answer is .161

Answer 4

Yes, trig substitution is the way to go (you would let \(x=2\tan\theta\) in this case).

Answer 5

and \[dx=2\sec^2(\theta) d \theta\]

Answer 6

Correct; Also, I would recommend changing the limits of integration here too (it simplifies the process a little bit).

Note that if \(\large x=0\), then \(\large 2\tan\theta = 0\implies \tan\theta= 0\implies\theta = 0\) and if \(\large x=2\), then \(\large 2\tan\theta = 2\implies\tan\theta =1\implies \theta=\dfrac{\pi}{4} \).

Thus, we see that if \(\large 0\leq x\leq 2\) (i.e. old limits of integration), then \(\large 0\leq \theta\leq\dfrac{\pi}{4} \) (i.e. new limits of integration).

Does this make sense? :-)

Answer 7

awesome! That makes a lot of sense...especially with the answers I was given (it was a multiple choice question)

Answer 8

and should I go ahead and expand the denominator? or just leave it?

Answer 9

Since you're making the substitution x=2tanθ, then
\[\large (x^2+4)^2=(4\tan^2\theta+4)^2=(4(\tan^2\theta+1))^2=(4\sec^2\theta)^2=16\sec^4\theta\]So after putting all the pieces together, what integral do you get in the end (before actually evaluating it)?

Answer 10

\[4\int\limits_{0}^{\pi/4}\frac{ 1 }{ \sec^2\theta }\]

Answer 11

I had \[\frac{ 4\sec^2\theta }{ 16\sec^4\theta }\] and simplified it

Answer 12

and wouldnt the integral of 1/sec^2(theta) be 1/tan(theta)?

Answer 13

no, I guess not. Per the integral calculator

Answer 14

my person left :(

Answer 15

Sorry, I was working on another problem (took a while). Note that
\(\large \dfrac{1}{\sec^2\theta} = \cos^2\theta = \dfrac{1+\cos(2\theta)}{2}\).

So after making the trig substitution, you should see that
\[\large \begin{aligned}16\int_0^2 \frac{1}{(x^2+4)^2}\,dx \xrightarrow{x=2\tan\theta}{} &\phantom{=} 16\int_0^{\pi/4} \frac{1}{4\sec^2\theta}\,d\theta\\ &= \frac{1}{4}\int_0^{\pi/4}\cos^2\theta\,d\theta\\ &= \frac{1}{8}\int_0^{\pi/4}1+\cos(2\theta)\,d\theta\end{aligned}\]
Does this make sense? Do you think you can take things from here? :-)

Answer 16

Ahh, okay. Cool. I wasn't sure if I should go in that direction or tangent. But yeah, I'm sure I can get it from here. Thank you for your help! I'm studying for my cal 2 final and we did trig sub like...3 months ago, lol

Answer 17

Fixing a couple typos, this is what I should have written originally:
\[\large \begin{aligned}16\int_0^2 \frac{1}{(x^2+4)^2}\,dx \xrightarrow{x=2\tan\theta}{} &\phantom{=} 16\int_0^{\pi/4} \frac{1}{4\sec^2\theta}\,d\theta\\ &= \color{red}{4}\int_0^{\pi/4}\cos^2\theta\,d\theta\\ &= \color{red}{2}\int_0^{\pi/4}1+\cos(2\theta)\,d\theta\end{aligned}\]
That looks much better now. Hopefully my tiny mistakes didn't confuse you! xD

Answer 18

awesome, I'll look over it and work it a few times. Thank you very much :)