Question

Evaluate the integral, check my work ( I messed up somewhere).

Answer 1

\[\int\limits_{}^{}\frac{ 9xlnx }{ \sqrt{x^2-16} }dx\]

Answer 2

\[x=4\sec \theta, dx = 4\sec \theta \tan \theta d \theta \]\[=9\int\limits_{}^{}\frac{ 4\sec \theta \ln(4 \sec \theta ) }{ \sqrt{16\tan^2 \theta} }(4\sec \theta \tan \theta d \theta)\]\[=9\int\limits_{}^{}\frac{ 4\sec^2\theta \ln(4\sec \theta)4\tan \theta }{ 4\tan \theta }=36\int\limits_{}^{} \sec^2\theta \ln(4\sec \theta)\]
Integration by Parts:
\[u = \ln(4\sec \theta), du = \frac{ 1 }{ 4\sec \theta }(4\sec \theta \tan \theta d \theta )\]\[dv =\sec^2\theta d \theta, v = \tan \theta\]\[IBP =\ln(4\sec \theta)\tan \theta - \int\limits_{}^{}\tan^2\theta d \theta \]\[=\ln(4\sec \theta)\tan \theta - \int\limits_{}^{}(\sec^2 \theta-1)d \theta \]\[=\ln(4\sec \theta)\tan \theta - \tan \theta + \theta \]
\[\frac{ x }{ 4 }=\sec \theta, \theta = 4\sec^{-1} \theta\]\[=36[\ln(x)\frac{ \sqrt{x^2-4} }{ 4 }-\frac{ \sqrt{x^2-4} }{ 4 }+4\sec^{-1}(x)]+C\]

Answer 3

Hmm I think Parts right from the start would've been easier :O
But anyway, looking it over..

Answer 4

\[\frac{x}{4}=\sec(\theta) \implies \theta=\sec^{-1}(\frac{x}{4})\]

Answer 5

\[\theta \neq 4 \sec^{-1}(\theta)\]

Answer 6

Ah, I see. Did I do tan theta correctly?

Answer 7

checking...
I think you forgot to square 4
\[\sec(\theta)=\frac{x}{4} (=\frac{hyp}{adj})\]