Question

integral of sqrt(36+x^2)dx

Can anyone help me with this?
I used trig identities and somehow got to
integral of 36 sec^3(θ) dθ but I'm not sure if this is right or what my next step is..

Answer 1

What is an integral?

Answer 2

Use a substitution \(x=6\tan(z)\), the integral will be easy with this substitution. After you evaluate the integral in terms of \(z\), substitute back for \(x\).

Answer 3

I did that and got
\[36\sqrt{1+\tan^2\theta} (\sec^2\theta) d \theta\]
but I'm not sure where to go from here.
Another substitution?

Answer 4

That expression can be simplified into \(36\sec^3\theta\).
To evaluate this integral, use integration by parts with \(u=\sec\theta \) and \(dv=\sec^2 xdx\).

Answer 5

Try to do it and let me know if you're stuck at any point.

Answer 6

okay so I have this
\[36(\sec \theta \tan \theta - \int\limits_{}^{} \tan^2 \theta \sec \theta d \theta\]

but I'm stuck at the second part of the integral
I made tan^(θ) = 1-sec^2(θ) and set u = secθ, dv = 1-sec^2(θ)
Is this correct?

Answer 7

Let our integral be \(I\). Then by integration by parts
\[I=\sec\theta \tan\theta -\int \tan^2\theta \sec\theta d\theta=\sec\theta\tan\theta-\int\sec^3\theta d\theta+\int \sec\theta d\theta. \]
So, \(I=\sec\theta\tan\theta-I+\int \sec\theta d\theta \implies \large { 2I=\sec\theta \tan\theta+\int\sec\theta d\theta}. \)
Evaluate \(\int\sec\theta\) and solve for \(\bf I\).

Answer 8

Don't forget to multiply by \(36\) at the end, since the original integral is \(36 I\).

Answer 9

I got it now, thanks :)