Question

Calculus II: Integrate x/sqrt (225x^2 - 1) using trig substitution.

Answer 1

which trig substitution has the formula \[\sqrt{x^2-a^2}\]?

Answer 2

it's tan^2 u = sec^2 u - 1

Answer 3

But can we use it in this one since the constant (or w/e it's called) is 1 ?

Answer 4

\[x=asec(\theta)\]) which then put into the integral for the substitution would be \[\tan^2(\theta)\] and you can use this even though there is a constant.

Answer 5

pull the 225 out of the main equation so you get \[\sqrt{225(x^2-(1/225)}\]

Answer 6

225 is 15 squared.

Answer 7

so i think x= (1/15)sec(theta)

Answer 8

Does that make sense? I'm not sure if I am explaining it correctly or easily.

Answer 9

I am trying to do it using your way

Answer 10

Anyways i think you get after you square root everything you should get \[(x)/15\tan(\theta)\] and \[x= (1/15)\sec(\theta)\]] . So it should look like this: \[((1/15)\sec(\theta))/(15\tan(\theta))\]

Answer 11

pull out the constants so (1/15)/15 which i think gives you one, unless my math is wrong, BE SURE TO DOUBLE CHECK ME!!!!

Answer 12

so you now have \[\sec(\theta)/\tan(\theta)\] sec is 1/cos and tangent is sin/cos so you have technically speaking (1/cos)/(sin/cos) which is (1/cos)*(cos/sin) so the cosine's cancel leaving you with 1/sin which is \[\csc(\theta)\] integrate.

Answer 13

the antiderivative of \[\csc(\theta)\] is \[\ln \left| \csc(\theta)-\cot(\theta) \right| +C\] plug in for theta which since x=(1/15)sec(theta) solve for theta.

Answer 14

so arcsec(15x)= theta

Answer 15

so lnl csc(arcsec15x)- cot(arcsec15x)+C....I do not know how to simply it from there, sorry but at least I got you an answer :D!

Answer 16

Hopefully that made sense and helped!!!

Answer 17

Thank you for helping me out. :D

Answer 18

No problem!!! I just did this on a test for my calc 2 class so hopefully my math was right and I didn't confuse you!!

Answer 19

You kinda did confuse me. LOL. But I figured it out.

Answer 20

Sorry, but at least you figured it out!!

Answer 21

I think you made a mistake somewhere because I got this answer and it is correct according to the book. \[\frac{ 1 }{ 225 }\sqrt{225x ^{2}-1}\]

After you pull out the constants, you get \[\frac{ 1 }{ 225 } \int\limits_{}^{} \frac{ \sec \theta \sec \theta \tan \theta }{ \tan \theta} d\theta \]