Hi, I'm Shaun! I'm currently in my third year of studying for a masters in Mathematics at the University of Exeter.

Hi, I'm Shaun! I'm currently in my third year of studying for a masters in Mathematics at the University of Exeter.

I believe that it is very important in mathematics to understand what you are taught, not just blindly learn formulae! As such, I focus on teaching the ideas behind the method, not just the method itself. I also think it's very important to practise questions yourself, so the sessions are interactive. The actual structure of the sessions is up to you! I can go through exam papers or focus on particular topics you struggle with. If you have any questions, feel free to contact me. You can also arrange a free 'Meet the Tutor' session through this website to get to know me a little before you decide whether to go ahead with some sessions.

I believe that it is very important in mathematics to understand what you are taught, not just blindly learn formulae! As such, I focus on teaching the ideas behind the method, not just the method itself. I also think it's very important to practise questions yourself, so the sessions are interactive. The actual structure of the sessions is up to you! I can go through exam papers or focus on particular topics you struggle with. If you have any questions, feel free to contact me. You can also arrange a free 'Meet the Tutor' session through this website to get to know me a little before you decide whether to go ahead with some sessions.

No DBS Check

**Find the following integral: ****∫ x sin(x) dx**

This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.

Recall that if you have an integral of the form

∫ u(dv/dx) dx

it can be written as

uv – ∫ v(du/dx) dx.

We need to decide which part we will differentiate (as in, which part is *u*), and which part we will integrate (as in, which part is *dv/dx*).

We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.

From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let

u = x, which implies du/dx = 1

and let

dv/dx = sin(x). Integrating this to get v gives v = –cos(x).

So our integral is now of the form required for integration by parts.

∫ x sin(x) dx

= ∫ u(dv/dx) dx

= uv – ∫ v(du/dx) dx

= –x cos(x) – ∫ –cos(x)*1 dx

= –x cos(x) – ∫ –cos(x) dx

= –x cos(x) + ∫ cos(x) dx

The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll call *c*. So the final answer is

∫ x sin(x) dx = –x cos(x) + sin(x) + c

**Find the following integral: ****∫ x sin(x) dx**

This question is a good candidate for the integration by parts method, as it is the product of two different 'parts'.

Recall that if you have an integral of the form

∫ u(dv/dx) dx

it can be written as

uv – ∫ v(du/dx) dx.

We need to decide which part we will differentiate (as in, which part is *u*), and which part we will integrate (as in, which part is *dv/dx*).

We can note that continuously differentiating sin(x) results in a loop of cos(x), –sin(x), –cos(x), sin(x)..., whereas differentiating x once gives 1.

From this, it seems to make sense that we would want to differentiate the x part (so u is x) and therefore integrate the sin(x) part (so dv/dx is sin(x) ). So, let

u = x, which implies du/dx = 1

and let

dv/dx = sin(x). Integrating this to get v gives v = –cos(x).

So our integral is now of the form required for integration by parts.

∫ x sin(x) dx

= ∫ u(dv/dx) dx

= uv – ∫ v(du/dx) dx

= –x cos(x) – ∫ –cos(x)*1 dx

= –x cos(x) – ∫ –cos(x) dx

= –x cos(x) + ∫ cos(x) dx

The integral of cos(x) is equal to sin(x). We can check this by differentiating sin(x), which does indeed give cos(x). Finally, as with all integration without limits, there must be a constant added, which I'll call *c*. So the final answer is

∫ x sin(x) dx = –x cos(x) + sin(x) + c