MHB Struggling with a trig indefinite integration

[iadg87]

So here is the problem: Find the anti-derivative of sec 3x(sec(3x) + tan(3x))

Now I have tried foiling it out, and I am stuck at the part where I need to anti-derive Sec(3x)Tan(3x).

Any help/tips would be greatly appreciated.

 

[Greg]

What's the derivative of sec(x)?

 

[Prove It]

So here is the problem: Find the anti-derivative of sec 3x(sec(3x) + tan(3x))

Now I have tried foiling it out, and I am stuck at the part where I need to anti-derive Sec(3x)Tan(3x).

Any help/tips would be greatly appreciated.

$\displaystyle \begin{align*} \int{ \sec{(3x)} \left[ \sec{(3x)} + \tan{(3x)} \right] \, \mathrm{d}x} &= \int{ \sec^2{(3x)} + \sec{(3x)}\tan{(3x)} \,\mathrm{d}x} \\ &= \frac{1}{3}\tan{(3x)} + \int{\sec{(3x)}\tan{(3x)}\,\mathrm{d}x } \\ &= \frac{1}{3}\tan{(3x)} + \int{ \frac{\sin{(3x)}}{\cos^2{(3x)}}\,\mathrm{d}x} \\ &= \frac{1}{3}\tan{(3x)} - \frac{1}{3} \int{ -3\sin{(3x)}\cos^{-2}{(3x)}\,\mathrm{d}x } \\ &= \frac{1}{3}\tan{(3x)} - \frac{1}{3} \int{ u^{-2}\,\mathrm{d}u} \textrm{ if we let } u = \cos{(3x)} \implies \mathrm{d}u = -3\sin{(3x)}\,\mathrm{d}x \\ &= \frac{1}{3}\tan{(3x)} - \frac{1}{3} \left( \frac{u^{-1}}{-1} \right) + C \\ &= \frac{1}{3}\tan{(3x)} + \frac{1}{3} \left[ \frac{1}{\cos{(3x)}} \right] + C \\ &= \frac{1}{3} \tan{(3x)} + \frac{1}{3}\sec{(3x)} + C \end{align*}$