∫e ax cos(bx) dx

∫eaxcos(bx) dx


1 Answers

Badiuddin askIITians.ismu Expert
148 Points
14 years ago

Dear tejas akole

let I=∫eaxcos(bx) dx

integrate by taking eax    as a first function

I=1/b eaxsinbx -a/b∫eaxsinbx dx

now againt integrate bt taking e^ax as first function

I=1/b eaxsinbx +a/b2 eaxcos(bx) -a2/b2∫eaxcos(bx) dx

I=1/b eaxsinbx +a/b2 eaxcos(bx) -Ia2/b2

I= eax/(a2+b2) {a cosbx +b sin bx} +c

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation.

 All the best tejas akole.
Askiitians Experts

Think You Can Provide A Better Answer ?


Get your questions answered by the expert for free