Currently unavailable: for new students

Degree: Mathematics (Masters) - Edinburgh University

I am a student of mathematics at the University of Edinburgh. I'm passionate about my subject and want to help you discover a love for maths too!

In my sessions we'll use extensive examples to explore the benefits and drawbacks of different mathematical methods, and to highlight areas where care should be taken. We won't just focus on calculation though - I'll try to always emphasise the origins of every formula we use, as this is what makes formulae easy to memorise. Learning will always be guided by you and I'll always take the time to first understand where you struggle to tailor the session to your needs: because everyone learns differently.

#### Subjects offered

SubjectQualificationPrices
Maths A Level £20 /hr
Maths GCSE £18 /hr

#### Qualifications

 CRB/DBS Standard No CRB/DBS Enhanced No

#### General Availability

Currently unavailable: for new students

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#### Ratings and reviews

5from 4 customer reviews

Justine (Parent) August 10 2016

great tutor

Justine (Parent) May 18 2016

Great lesson.

Josh (Student) May 17 2016

great helped a lot with my maths

Justine (Parent) August 8 2016

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### How do you integrate by parts?

To integrate the product of two functions - say f(x) = xsinx, the product of g(x) = x and h(x) = sin(x) - the we use integration by parts.  Since integration is the inverse of differentiation, we'll appeal to the Product Rule for differentiation to find a rule for integration. The Product Rul...

To integrate the product of two functions - say f(x) = xsinx, the product of g(x) = x and h(x) = sin(x) - the we use integration by parts.

Since integration is the inverse of differentiation, we'll appeal to the Product Rule for differentiation to find a rule for integration. The Product Rule says that (fg)' = f'g + g'f. Rearranging giives:

f'g = (fg)' - g'f.

Integrating we have:

Integral(f'g) = Integral((fg)') - Integral(g'f), or:

Integral(f'g) = fg - Integral(g'f).

This is integration by parts. We designate one of the functions to be f' and one to be g. You can see on the far right we have another integral! For the method to be of use, we need to make the new integral simpler (or at least no more complicated) than the one we began with. So, we want g'f to be simpler then f'g: in other words, when we assign functions f' and g we want g to be the one that gets simpler when differentiated. Let's try an example: let's try and integrate xsinx dx. First, we will choose g to equal x, since x becomes 1 under differentiation, but sinx goes to cosx.

I = Integral(xsinx) = xIntegral(sinx) - Integral(x'Integral(sinx))

I = -xcosx - Integral(-cosx)

I = sinx - xcosx +C

So as long as we choose f' and g with some forethought, it's easy enough to integrate products of functions. Note: Ususally we denote the rule with f' = dv and g = du, so

Integral u dv  = uv - Integral v du.

This is just a fancy way of stating what we've already established.

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