Question

Let I = ſeat cosbx dx​

Answer 1

The question is irrelevant I think

Answer 2

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