After using a basic trig identity, this integral becomes a simple case of integration by parts.
Note that x/cos^2(x)=x*1/cos^2(x)=xsec^2(x). That means we can rewrite the integral as intxsec^2(x)dx. Because x and sec^2(x) are being multiplied, we use integration by parts, which tells us:
intudv=uv-intvdu
In other words, we make a relatively complicated integral like intxsec^2(x)dx simpler. The only difficulty is choosing our u and dv. In this case, u=x and dv equals everything else, namely sec^2(x)dx (when dealing with algebra and trig functions in integration by parts, always choose the algebraic function as your u). Now we find du and v:
u=x=>(du)/dx=1=>du=dx
dv=sec^2(x)dx=>intdv=intsec^2(x)dx=>v=tan(x)
We can ignore the constant of integration in this case.
Now we know u=x, du=dx, v=tan(x). Plugging these in to the integration by parts formula, we have:
intxsec^2(x)dx=xtan(x)-inttan(x)dx
We could do inttan(x)dx, but that would be a waste of time because there is a far easier way: consult a table of integrals! After doing so, we find the result is either lnabs(sec(x)) or -lnabs(cos(x)) (both are the same thing). I'm going to use -lnabs(cos(x)).
intxsec^2(x)dx=xtan(x)-(-lnabs(cosx))+C
=xtan(x)+lnabs(cosx)+C
And that is our final answer.
