Integration by parts

∫2e^(x).Sin(x)Cos(x) dx

v= Sin(x)Cos(x)
dv/dx = Cos^2 (x) - Sin^2 (x) = Cos2(x)

du/dx = 2e^(x)
u= 2e^(x)

∫2e^(x).Sin(x)Cos(x) dx = 2e^(x) . Sin(x)Cos(x) - ∫ 2e^(x). Cos2(x)

v=Cos2(x)
dv/dx = -2sin2(x)

du/dx= 2e^(x)
u= 2e^(x)

∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [ 2e^(x).Cos2(x) - ∫2e^(x).-2sin2(x) dx]
=>
∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [2e^(x).Cos2(x) + 2∫2e^(x).sin2(x) dx]
=>
∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x) - 2∫2e^(x).sin2(x) dx

rearranging this whole mess
∫2e^(x).Sin(x)Cos(x) + 2∫2e^(x).sin2(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x)

3∫2e^(x).Sin(x)Cos(x) dx = e^(x).(sin2(x)-2Cos2(x)
textbook answer is (e^(x) /5).(sin2(x)-2Cos2(x)
where does the 5 come from? can anyone spot the mistake please

(edited 6 years ago)

Reply 1

MikeyA

do you still need an answer?

Reply 2

Carlos Nim

yes please

(edited 6 years ago)

Reply 3

MikeyA

have you got like a Snapchat or messenger because tsr isn't letting me upload a photo of the working out

Reply 4

Notnek

Original post by MikeyA

have you got like a Snapchat or messenger because tsr isn't letting me upload a photo of the working out

It would be better if you can find the OPs mistake if possible instead of posting your own working.

okee

To Op, I think it would be easier for you to simplify the integral in question, preferably using sin2x = 2sinxcosx so you can remove sinxcosx from your integral.

It would also help to keep your numbers outside the integral.

Reply 7

Notnek

Original post by Carlos Nim

Your mistake is going from here

∫2e^(x).Sin(x)Cos(x) + 2∫2e^(x).sin2(x) dx

to here

3∫2e^(x).Sin(x)Cos(x) dx

Please try again and if you get the same thing post your working in full for this step.

As the poster above said, it would have been much easier if you changed sin(x)cos(x) using the identity before doing any integration and you probably wouldn't have made this mistake.

Reply 8

Carlos Nim

∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [ 2e^(x).Cos2(x) - ∫2e^(x).-2sin2(x) dx]
=>
∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - [2e^(x).Cos2(x) + ∫4e^(x).sin2(x) dx]
=>
∫2e^(x).Sin(x)Cos(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x) - 4∫e^(x).sin2(x) dx]
=>
∫2e^(x).Sin(x)Cos(x) + 4∫e^(x).sin2(x) dx = 2e^(x).Sin(x)Cos(x) - 2e^(x).Cos2(x)
=>
6∫e^(x).Sin(x)Cos(x) dx = e^(x).(sin2(x)-2Cos2(x)
=>
∫e^(x).Sin(x)Cos(x) dx = e^(x)/6.(sin2(x)-2Cos2(x)

Im not sure about this, anyway i got 6 , the answer required is 5

(edited 6 years ago)

Reply 9

MikeyA

I would suggest to start again and simplify the integral first using the identity I gave earlier, it would make the simplification at the later stages much easier and this you would be less likely to mess up