What is the primitive of sinx/cos^2x?

[Krushnaraj Pandya]

Homework Statement

∫e^(-x)(1-tanx)secx dx
2. Attempt at a solution
I know ∫e^x(f(x)+f'(x))=e^x f(x)
and I intuitively know f(x) could be secx here and therefore f'(x) will be secxtanx but I can't figure out how to reach that

 

[BvU]

So you want a primitive of $\displaystyle { \sin x\over \cos^2 x}$ .

It is almost handed over on a silver platter: if $\ \ -\displaystyle {\sin x\over \cos^2 x}\$ is $f'(x)$, what do you have left over for $f(x)$ ?

 

[Krushnaraj Pandya]

So you want a primitive of $\displaystyle { \sin x\over \cos^2 x}$ .

It is almost handed over on a silver platter: if $\ \ -\displaystyle {\sin x\over \cos^2 x}\$ is $f'(x)$, what do you have left over for $f(x)$ ?

I know the derivative of -secx is -sinx/cos^2 x. the first trouble is that the problem has e^(-x) instead of e^(x). the second is that -sin/cos^2 is the derivative of -secx, not secx. I'm sure these two things tie together somehow through a basic simplification but I can't figure this basic simplification out

 

[Krushnaraj Pandya]

So you want a primitive of $\displaystyle { \sin x\over \cos^2 x}$ .

It is almost handed over on a silver platter: if $\ \ -\displaystyle {\sin x\over \cos^2 x}\$ is $f'(x)$, what do you have left over for $f(x)$ ?

AHHH! how INCREDIBLY stupid of me. I just had to put -x=t (sorry you had to read this question). I'm never going to be a scientist this way...