I'm very new for calculus.
please solve this problem. thank you very much.
I gave you a useless step, sorry i wasn't thinking about it.
Anyway i'm going to suggest something radical and an easy way out.
But i'm not sure if you are allowed to leave your answers in terms of non elementary fuctions.
Divide everything by cos^7 x
you get $\displaystyle \frac{tan^{7} x}{ 1 + tan^{7} x} = 1 - \frac{1}{1 + tan^{7}x} $
use a substitution u = tanx
du = sec^2 x = ( u^2 - 1)
you get
$\displaystyle \frac{1}{ (u^{2} -1)( 1+u^{7})}$ you can do this
$\displaystyle \frac{1}{(u-1)(u+1)(1+u^{7})}$
Partial fractions would be appropriate.
If you get something like
$\displaystyle \frac{Au+b}{(1+u^{7})}$
You may have to use the maclaurin series or do some more algrabra. I didn't try it so i'm not sure what you would end up with.
Good luck though. The algebra seems disgusting.
I popped it into an online integrator and it doesn't look pretty.
Wolfram Mathematica Online Integrator
I think you might get stuck with the partial fraction.
$\displaystyle \frac{1}{(u^{2}-1)(u^{7} +1)}$ =
$\displaystyle \frac{-3x^{4}+2x^{3} -4x^{2} +2x -3}{7( x^{6}-x^{5}+x^{4} -x^{3} +x^{2}-x +1)} + \frac{1}{4(x-1)}- \frac{1}{4(x+1)} - \frac{1}{14(x+1)^{2}}$
Did it with pc.
Forget about this integral; it's not the kind of thing you would see in an exam.