Question

Evaluate the Integral.

Answer 1

\[\int\limits_{}^{}\frac{ \sqrt{9+x^2} }{x }\]

Answer 2

trig substitution

Answer 3

I used trig substitution and got \[\int\limits_{}^{}\frac{ \sqrt{9+9\tan^2\theta} }{ 3\tan \theta } 3\sec^2\theta\]

Answer 4

Then I simplified to \[\int\limits_{}^{}\frac{ 3\sec \theta }{3\tan \theta } 3\sec^2\theta \]

Answer 5

\[3\int\frac{\sec^4\theta}{\tan\theta}\,\mathrm d\theta=3\int\frac{\dfrac{1}{\cos^4\theta}}{\dfrac{\sin\theta}{\cos\theta}}\,\mathrm d\theta=3\int\frac{\mathrm d\theta}{\sin\theta\cos^3\theta}=3\int\csc\theta\sec^3\theta\,\mathrm d\theta\]Any ideas how to continue from here?

Answer 6

Shouldn't it be sec to the third?

Answer 7

Sorry, I misread that as a \(3\) (poor eyesight). You're right, the final integral should be \(\displaystyle3\int\csc\theta\sec^2\theta\,\mathrm d\theta\).

My previous question still stands.

Answer 8

Would you do integration by parts?

Answer 9

Not a bad idea (probably, I haven't checked). What kind of \(u\)-\(v\) setup do you propose?

Answer 10

I have no idea. Im not sure it will work now.

Answer 11

You could try something like
\[\begin{matrix}
u=\csc\theta&\mathrm dv=\sec^2\theta\,\mathrm d\theta\\[1ex]
\mathrm du=-\csc\theta\cot\theta\,\mathrm d\theta&v=\tan \theta
\end{matrix}\]so the integral becomes
\[\int\csc\theta\sec^2\theta\,\mathrm d\theta=\csc\theta\tan\theta+\int\csc\theta\,\mathrm d\theta\]which is doable.

Answer 12

Alternatively, you can play around with partial fractions. Rewrite the integrand in terms of sines:
\[\csc\theta\sec^2\theta=\frac{1}{\sin\theta\cos^2\theta}=\frac{1}{\sin\theta(1-\sin\theta)(1+\sin\theta)}\]which decomposes into
\[\frac{1}{\sin t}+\frac{1}{2(1-\sin t)}-\frac{1}{2(1+\sin t)}\]

Answer 13

\(\theta\), not \(t\)...

Answer 14

So would it end up being \[\csc \theta \tan \theta+\ln |\csc \theta + \cot \theta| +C\]

Answer 15

I believe the log term should be negative.

Answer 16

Also, you can simplify the first term: \(\csc\theta\tan\theta=\dfrac{1}{\sin\theta}\times\dfrac{\sin\theta}{\cos\theta}=\sec\theta\).

Answer 17

Oh yes you're right.

Answer 18

What would I do now that I have \[\sec(\tan^{-1} (\frac{ x }{ 3 }))-\ln|\csc(\tan^{-1} (\frac{ x }{ 3 })+\cot(\tan^{-1} (\frac{ x }{ 3})| \]