Ask your own question, for FREE!
Calculus1 7 Online
OpenStudy (anonymous):

Just need someone to check my answer on an integration problem ... Integral of 1/(1+4x^2) dx = 1/4 arctan(4x)+C? Or am I completely off?

zepdrix (zepdrix):

\[\large \int\limits \frac{1}{1+4x^2}dx\] Hmm I think you're very close. You just forgot that your 4's get squared when you make a substitution. So you don't want to be substituting 1/4. You want a 1/2. (Assuming that's the route you took toward solving this). \[\large x=\frac{1}{2}\tan \theta\]\[\large \int\limits\limits\limits \frac{1}{1+4x^2}dx \qquad = \qquad \int\limits\limits\limits \frac{1}{1+4(\frac{1}{2}\tan \theta)^2}dx \qquad = \qquad \int\limits\limits\limits \frac{1}{1+\tan^2 \theta}dx\]

zepdrix (zepdrix):

\[\large \color{royalblue}{dx=\frac{1}{2}\sec^2\theta \;d \theta}\] \[\large \int\limits\frac{1}{1+\tan^2 \theta}\color{royalblue}{dx}\]

OpenStudy (anonymous):

I'm a little confused about the substitution part. How did you get that x = 1/2 tan x?

zepdrix (zepdrix):

We see a form that looks similar to this in the denominator.\[\large a^2+x^2\] When we see this form, we make the substitution, \[\large x=a \tan \theta\]\[\large a^2+x^2 \qquad \rightarrow \qquad a^2+(a \tan \theta)^2 \qquad \rightarrow \qquad a^2(1+ \tan^2 \theta)\]Using our trig identity, this will simplify to,\[\large a^2(\sec^2 \theta)\] In the problem we have here, it appears our a value not where we would like it to be. We have something looking a little more like this,\[\large 1+a^2x^2\] We will make the substitution, \[\large x=\frac{1}{a}\tan \theta\]That will get rid of the a value for us.\[\large 1+a^2x^2 \qquad \rightarrow \qquad 1+a^2\left(\frac{1}{a}\tan \theta\right)^2 \qquad \rightarrow \qquad 1+\tan^2 \theta\]\[\large = \sec^2\theta\]

zepdrix (zepdrix):

If you've learned this using a different method, such as based off of memorization, then this might be a lil confusing D:

zepdrix (zepdrix):

Our goal is to get rid of the addition or subtraction that may be in the denominator by using a Trig Identity. It will make the integral much much easier to deal with.

OpenStudy (anonymous):

no i appreciate your explanation! I have to run now, but I think I can make sense of this now. :) If I get terribly confused trying to understand it, I'll mention you in a reply later on. :D

OpenStudy (anonymous):

yeah that makes sense :)

zepdrix (zepdrix):

kk c: have nice day

OpenStudy (anonymous):

you too! Thanks!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!