Currently unavailable: for regular students
Degree: Maths and Philosophy (Bachelors) - Durham University
Hi, I'm Dan and I am in my third year of studying Maths and Philosophy at Durham University. I have a real interest in my subjects and hope that this level of understanding and enthusiam comes across in my tutorials.
I also have teaching experience, having assisted in lessons at GCSE and A Level, as well as more informal sessions among friends.
Outside of my studies, I love playing piano, going to live gigs and cooking.
About the Tutorials
With these sessions, you will be the one who decides what we cover. I look to address any issues you may have with your studies, and leave you with both the understanding and problem-solving ability to go out and tackle future problems yourself.
I also understand that learning is not something that happens in hour-long sessions, but something that takes time. Because of this I'm always happy for my students to message me whenever you need help.
I look forward to meeting you!
|Further Mathematics||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|Philosophy and Ethics||A-Level||A|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Bilal (Student) January 20 2016
Matthew (Parent) May 11 2015
Matthew (Parent) May 1 2015
Matthew (Parent) April 22 2015
The key to solving any integral of this form is to use the cosine rule:
cos(2x) = cos2(x) - sin2(x) = 2cos2(x) - 1 = 1 - 2sin2(x)
All of these forms are really helpful when solving problems such as this, and it's great if you can remmeber them, though if you get stuck in an exam, they can all be derived from the addition formulae that are probably on your fomula sheet!
So, using the above idenities, we know that:
2cos2(x) - 1 = cos(2x)
2cos2(x) = cos(2x) + 1
cos2(x) = (cos(2x) + 1)/2
So instead, we perform the integral of (cos(2x) + 1)/2, which we already know how to do.
=> (sin(2x))/4 + x/2see more
The integral of log(x) is not necessarily straight-forward. Though we can use the fact that d/dx(log(x)) = 1/x to help us.
Rather than simply trying to integrate log(x), we can use integration by parts on 1 x log(x) (as in 'one times' log(x)).
So we can differentiate the log(x) part and integrate the 1 part to give:
xlog(x) - ∫ 1 dx = xlog(x) - x
Note: if the middle step isn't clear, we can write it more explicitly as
u = log(x) v' = 1
u' = 1/x v = x
Where the rule for integration by parts is written as:
uv' = uv - ∫ u'v , where u and v are functions of xsee more