x Turn on thread page Beta
 You are Here: Home >< Maths

# Integration. XD C4 watch

1. Hi,

Is this correct?

Find the integral of xcos2x dx:

xsin2x / 2 - cos2x / 4 + c

Use the result above to find x cos^2(x) dx:

So far I got:

integral x [2cos^2(x) - 1] dx

integral 2x cos^2(x) - x dx

2integral x cos^2(x) - integral x dx...

Is this correct atm, also where do I go from here?
Thanks!
2. The identity cos(2A) = 2cos^2(A) - 1, rearrange gives cos^2(A) = (1 + cos(2A))/2

So integral of xcos^2(x) = integral of x*(1 + cos(2x))/2
so you can use your previous result.
3. (Original post by vc94)
The identity cos(2A) = 2cos^2(A) - 1, rearrange gives cos^2(A) = (1 + cos(2A))/2

So integral of xcos^2(x) = integral of x*(1 + cos(2x))/2
so you can use your previous result.
Thank you very much. Ive been seeing this a lot where part b is the same answer as part a. You just have to find out how to get there in relation with the previous part.

Turn on thread page Beta
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: April 8, 2011
Today on TSR

### Boyfriend slept with someone else

...we were on a break

Poll
Useful resources

## Make your revision easier

### 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

Can you help? Study help unanswered threads

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE