Yes, but I'd rather I helped YOU solve it.
Try a substitution [itex]u=tan(\theta), du=sec^2(\theta) d\theta[/itex]. It might not make this any easier, but it's the best method I can think of right now.
i tried it. but i didn't can
Well, you should get as your equation...
[tex]\int sec^3(\theta) d\theta[/tex]
And from there it's a simple integration by parts, taking u=sec(\theta) and dv=sec^2(\theta).
Moderator's note: thread moved from Calculus & Analysis
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