You are Here: Home >< Maths

# Integration by parts watch

1. ∫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)
where does the 5 come from? can anyone spot the mistake please
2. do you still need an answer?
4. have you got like a Snapchat or messenger because tsr isn't letting me upload a photo of the working out
5. (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.
6. okee
7. 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.
8. (Original post by Carlos Nim)
x
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.
9. ∫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)

10. 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

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 14, 2018
Today on TSR

### University open days

• University of Exeter
Wed, 24 Oct '18
Wed, 24 Oct '18
• Northumbria University
Wed, 24 Oct '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams