Question

Is the int of tsec^(2)2tdt = (t/2)tan(2t) - (1/4)log(sec2t) ?

Answer 1

Series of explanation of the integral at x = 0
{"(2 t^3)/3+(2 t^3 x^2)/3+(4 t^3 x^4)/9+(34 t^3 x^6)/135+(124 t^3 \
x^8)/945+(2764 t^3 x^10)/42525+O(x^11)"}

Answer 2

What? So \[\int\limits_{}^{}t \sec ^{2}2t\] =
{"(2 t^3)/3+(2 t^3 x^2)/3+(4 t^3 x^4)/9+(34 t^3 x^6)/135+(124 t^3 \ x^8)/945+(2764 t^3 x^10)/42525+O(x^11)"} ???

Answer 3

Hint: use integration by parts.

Answer 4

I did...and the answer i got is \[(t/2) \tan2t - (1/4)logsec2t + C\]

Answer 5

That's correct!
If you are not sure, differentiate it with respect to t, you'll get back t*sec^2(2t).
The other solution is a series expansion solution, which is also valid in cases where it is difficult to integrate analytically.

Answer 6

u = t
du= dt
dv = sec^(2)t
v = (1/2) tan2t

Answer 7

I see...thank you!

Answer 8

You're welcome! :)