Integration by parts

  • #1
Krushnaraj Pandya
Gold Member
697
73

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
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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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)## ?
 
  • #3
Krushnaraj Pandya
Gold Member
697
73
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
 
  • #4
Krushnaraj Pandya
Gold Member
697
73
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...
 

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