If limits considered from 0 to (pi/4) and without limits also.
I used maple which it gives:
(1/2)*sin(x)/cos(x)^2+(1/2)*ln(sec(x)+tan(x))
I used maple which it gives:
(1/2)*sin(x)/cos(x)^2+(1/2)*ln(sec(x)+tan(x))
wolphramalpha.com gives you the answer and a lot more (free). Try it! Janos
Si estamos de acuerdo con la efectividad de la integracion por partes. Al final llegas a tener una integral CÍCLICA, es decir llegas a la misma integral de Secante al cubo. ...Dios les bendiga a todos.
It is easy. I am bemused to find so many different answers! Differentiate (( sec x tan x + log ( sec x + tan x)) / 2) with respect to x, and you will get sec cubed x. See the attachment to find how to arrive at the answer.
My answer from wolfralpha.com is in the file attachment. Your sincerely; Anna Tomova
The integral value is:
(1/2)sin(x)/(cos(x))^2+(1/2)ln[sec(x)+tan(x)].
If limits considered from 0 to (pi/4), then we obtain
integral value=1.14779
I used Mathematica:
Integrate[(Sec[x])^3, x]
1/2 (-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x] Tan[x])
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