Question

Calculus II: Is it possible to integrate this question using trig substitution? integral x^3/sqrt(x^2+144) dx

Answer 1

=Q
Q

Answer 2

sorry i didn't mean to post that

Answer 3

it's all good

Answer 4

tangent*

Answer 5

It's been a long day -- use
\[ x = 12 \tan(u) \]
\[ dx = 12 \sec^2(u) du\]

Answer 6

\[\int\frac{x^3}{\sqrt{x^2+144}}dx\]
Using @Jemurray3's suggestion,
\[\int\frac{(12 \tan u)^3}{\sqrt{(12\tan u)^2+144}}\cdot (12 \sec^2 u \;du)\\
12^4\int\frac{\tan^3u\sec^2u}{\sqrt{144\tan^2 u+144}}du\\
12^3\int\frac{\tan^3u\sec^2u}{\sqrt{\tan^2 u+1}}du\]
I'm sure you can take it from here.

Answer 7

I got till \[12^{3}\int\limits_{}^{} \sec ^{3}u \tan u - \tan u \sec (u) du\] Now what to do next?

Answer 8

Integration by parts?

Answer 9

Rewrite that as ( something in here ) * sec(u)tan(u) du

Answer 10

That would be sec^2 (u) - 1

Answer 11

I think I got it.