Currently unavailable: for regular students
Degree: Mathematics & Statistics (Bachelors) - Warwick University
I am currently studying at the University of Warwick in my second year reading Mathematics and Statistics, completing my 1st year with a first.
Throughout secondary school, maths was always my favourite lesson not only because I excelled in it but also because of the satisfaction of solving difficult problems. I have always enjoyed explaining my thoughts behind a concept or providing guidance on methods I have found successful in tackling questions.I have experience teaching and handling children as I often volunteered with my fencing club and younger classes learning the tabla (musical instrument) while at school. I am very patient and am willing to find alternate ways of thinking about a problem to make understanding the mathematics behind it easier.
Maths is a subject where understanding combined with practice are the key to success. During sessions it is important that you tell me what you are struggling with and which topics/areas you find difficult. I will hopefully then be able to provide you with a sound understanding of the topic by using both explanations of concepts and proofs.
Please feel free to contact me if you have any questions or concerns or you can book a free meet the tutor session on this page. I look forward to meeting you!
|Maths||A Level||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|Economics & Business||A-Level||A|
|Before 12pm||12pm - 5pm||After 5pm|
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1. ʃ sinx*ln(cosx) dx
First notice the composition ln(cosx). To make the expression easier to integrate we try substitution (substitution is often useful when trying to integrate expressions that are or include compositions of functions).
2. So we let t=cosx.
Now we must find out what dx is in terms of dt. This is quite simple by just differentiating the substitution.
3. dt/dx=-sinx <=> dx=(-1/sinx)dt
Now we substitute everything into our integral from 1.
4. ʃ sinx*ln(t)*(-1/sinx) dt
The sinx from the top and bottom cancel.
5. ʃ -1*ln(t) dt
I have written a -1 since we will have to use integration by parts to integrate ln(t) with respect to t. (Reminder: ʃ u(dv/dx) dx = uv - ʃ v(du/dx) dx)
6. Let u=ln(t) and dv/dt=-1
Now we need to differentiate u to find du/dt and integrate dv/dt to find t.
7. du/dt=1/t and v=-t
Use the intergation by parts formula and that uv=-tln(t) and v(du/dt)=-t(1/t)=-1.
8. ʃ -1*ln(t) dt = -tln(t) - ʃ -1 dt = -tln(t)+t+C
! Remember the constant of integration and that because we used a substitution we must give our answer back in terms of x.
9. So using 2. again we get cosx(1-ln(cosx))+C as the answer.see more