How to integrate (e^t)*tan^2(t)

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Homework Statement



I am in the process of solving second order differential equation by the variation of parameters method. I need to calculate ∫(et)*(tant)2 dt

Homework Equations


(tanx)2 = (secx)2-1
∫(secx)dx= lnlsecx*tanxl

The Attempt at a Solution


I have used the trig identity to get this: ∫(et*((sect)2-1)dt
= ∫et*(sect)2 dt -∫(et) dt
=∫et*(sect)2 dt -et
but am stuck here. Any help would be greatly appreciated. Thanks!
 
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The problem explicitly asks to solve using variation of parameters. Perhaps I made a mistake in the algebra somewhere before this step; I will go to my professor. Thanks for your input, now I won't waste a hour or two staring at and trying to solve this integral.
 
I doubt it; I probably made a mistake. Here is the problem:

Find the general solution by variation of parameters:
y'' + y = (tanx)2

For these, I know you find the solution by finding and adding together the solution to the homogeneous equation with that of your particular solution.

I found the solution to the homogeneous equation to be:
yh = C1*e-t+C2

Then I needed to find the particular solution which I knew would be in the form u1*y1 + u2*y2
where
u1 = -∫ [itex]\frac{y<sub>2</sub>*g}{W[y<sub>1</sub>,y<sub>2</sub>]}[/itex] dt
and
u2= ∫ [itex]\frac{y<sub>1</sub>*g}{W[y<sub>1</sub>,y<sub>2</sub> ]}[/itex] dt

(g is (tanx)2, and W is the Wronskian)
 
Ah! I see. I was using reduction of power method and messed up. I had gotten r2 + r =0 whereas it should be r2+1=0 . Therefore I should get an imaginary root, and the solution is yh= C1*sint + C2*cost