Let I = ſeat cosbx dx
Answers
Answer:
To do this integral there are two main approaches that might be of interest. If you know about complex numbers have a look have a look at method I
Method I :
Let
B=∫eaxsin(bx) dx
Let
u=eax and dvdx=sin(bx)⟹dudv=aeax and v=−1bcos(bx)(6)
Now apply the integration by parts formula with (6)
∫udvdx dx=uv−∫vdudx dx∫udvdx dx=uv−∫vdudx dx
and
B=∫eaxsin(bx) dx=−1beaxcos(bx)+ab∫eaxcos(bx) dx=−1beaxcos(bx)+abA(7)
where
A=∫eaxcos(bx) dxA=∫eaxcos(bx) dx
Now apply integration by parts to AA to get
A=∫eaxcos(bx) dx=1beaxsin(bx)−ab∫eaxsin(bx) dx=1beaxsin(bx)−abB(8)
Now from (7) and (8) we have two equations in AA and BB (the integrals we defined as (1) and (2)):
B=−1beaxcos(bx)+abAB=−1beaxcos(bx)+abA
A=1beaxsin(bx)−abBA=1beaxsin(bx)−abB
Rearrange these into a pair of simultaneous equation in A and B(2)):
B=−1beaxcos(bx)+abAB=−1beaxcos(bx)+abA
A=1beaxsin(bx)−abBA=1beaxsin(bx)−abB
Rearrange these into a pair of simultaneous equation in AA and B
B=−1beaxcos(bx)+abAB=−1beaxcos(bx)+abA
A=1beaxsin(bx)−abBA=1beaxsin(bx)−abB
Rearrange these into a pair of simultaneous equation in AA and B
A+abB−abA+B=1beaxsin(bx)=−abeaxcos(bx)(9)(10)(9)A+abB=1beaxsin(bx)(10)−abA+B=−abeaxcos(bx)
Solve (9) and (10) as a pair of simultaneous equations to get
AB=eax(acos(bx)+bsin(bx))a2+b2=eax(−bcos(bx)+asin(bx))a2+b2.
I hope it is very helpful for you
Thank you