My son was blown away by his first tutorial with Jamie. He loved Jamie's enthusiasm and said that everything was explained so well that concepts he's found difficult for a term and a half at school were quickly made clear. Also really liked the way that Jamie kept testing his understanding with questions. Fantastic, many thanks.

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the session was extremely useful as everything was explained to me in a detailed and organised way. I was provided with different methods to do some problem solving which made things a lot easier and I was given some good advise that will definitely help in the future. Another important point, is that the lesson was very fun which made things much nicer as well.

Laleh, Parent from Buckinghamshire

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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

As soon as you see a question asking you to integrate the square of sin, cos or tan, your first approach should be to use trigonometric identities and double angle formulas.

For sin^{2}(X), we will use the **cos** double angle formula:

cos(2X) = 1 - 2sin^{2}(X)

The above formula can be rearranged to make **sin ^{2}(X)** the subject:

sin^{2}(X) = 1/2(1 - cos(2X))

You can now rewrite the integration:

∫sin^{2}(X)dX = ∫1/2(1 - cos(2X))dX

Because 1/2 is a constant, we can remove it from the integration to make the calculation simpler. We are now integrating:

1/2 x ∫(1 - cos(2X)) dX = 1/2 x (X - 1/2sin(2X)) + C

It is very important that as this is not a definite integral, we must add the constant C at the end of the integration.

Simplifying the above equation gives us a final answer:

∫sin^{2}(X) dX = **1/2X - 1/4sin(2X) + C**

Answered by Kyna F.

Studies Materials Engineering at Exeter

When integrating a long chain of functions, we can integrate each term seperately and combine them. Let us now integrate:

∫21x^{6}dx = 21∫x^{6}dx. Using the Power Rule [∫x^{a}dx = (x^{a+1}/a+1)], we can say that 21∫x^{6}dx = (21x^{7})/7 = 3x^{7}.

∫e^{2x}dx. Now let u = 2x. du/dx = 2 so dx = du/2. Substitute both in to get:

∫(e^{u}/2)du = 1/2∫e^{u}du. This is a common integral, which gives us 1/2 e^{u} = 1/2 e^{2x}.

∫(1/x)dx. This is a common integral which equals ln |x|

∫6dx = 6∫dx = 6x (Integration of an integer).

We then combine all the terms to give us 3x^{7} - e^{2x}/2 - ln |x| + 6x.

When ever we integrate without limits, we have to add a constant c. This is unknown, unless addition information is given, so we call this C. Hence, the answer is:

3x^{7} - e^{2x}/2 - ln |x| + 6x + C

Write as y=uv u=x v=e^{x}

du/dx=1 dv/dx=e^{x}

Using the product rule, dy/dx=v*du/dx + u*dv/dx

So dy/dx=e^{x}(1)+x(e^{x})=e^{x}(1+x)

Answered by Isabel R.

Studies Mathematics at Manchester

So we have the equation initially in the form 3^(5x-2)=4^(6-x), and as the solution involves log10, then a sensible first move would be to take log10 of both sides, giving log10(3^(5x-2)) = log10(4^(6-x)).

Using the log law that loga(k) = kloga, we can rewrite this as (5x-2)log10(3) = (6-x)log10(4).

Then expanding the brackets, and taking the x values to one side and the constants to the other gives, 5xlog10(3) + xlog10(4) = 6log10(4) + 2log10(3).

Taking x out as a factor, we get x(5log10(3) + log10(4)) = 6log10(4) + 2log10(3).

Using the log law we made use of earlier in the question (loga(k) = kloga), this can be written as x(log10(243) + log10(4)) = log10(4096) + log10(9).

We can then use the log law, loga(b) + loga(c) = loga(bc), to write the equations as x(log10(972)) = log10(36864).

Now all we have to do is divide by log10(972), and we have our answer, so,

x = log10(36864) / log10(972).

Answered by Eloise B.

Studies Mathematics at Sheffield

Tricky. Definitely can't do it by inspection (we don't know any fuction that just differentaties to ln(x)), it's not like we've really got anything to substitute u for if we wanted to do it by substitution and integration by parts requires two different terms being multiplied together; and we only have one! It seems like all of the ways we know how to integrate aren't going to help us much??

Maybe one of them could though. If we rewrite ln(x) = 1*ln(x) we at least have two terms in order to do integrate by parts. choosing which is u and which is dV/dx isn't going to be very hard; if we took ln(x) = dV/dx then we'd have to integrate it immediately, which was the whole problem! u = ln(x) it is then.

this gives us du/dx = 1/x and V = the integral of 1 dx = x

This looks promising. Our formula for integration by parts ( derived from the product rule) is:

Integral(u*dV/dx) = uV - Integral(du/dx*V) Substiting the values we just got into this gets us:

Integral(ln(x)) = xln(x) - Integral(x*1/x) Awesome. x*1/x = 1 and we can definitely integrate that;

Answer = xlnx - Integral(1) = xlnx - x = x(lnx-1) + c [try not to forget the plus c!]

There you go. A little bit of creativity required, and we turned a seeming dead end into a complete solution! This result could definitely be useful when we're integrating more complex functions, like x/ln(x).

The first thing we notice is that this differential equation is seperable, meaning we can get all of our y's on the left with a dy and all of our x's on the right with a dx. Doing this by multiplying both sides by e^2y and dx gives:

e^2y dy = x√(x^2+3) dx

Now we need to integrate on both sides. Integrating the left with respect to y won't be too bad, we'll get

e^2y/2 Whenever we integrate by inspection, it's a good idea to check that we have the right answer by differentiating and making sure we get back what we put in to start with. This works in this case, as differentiating by the chain rule with u = 2y gets us 2(e^2y/2) = e^2y, which is what's on the left.

So far so good, but a tricky one now. On the right we have an x-term multiplied by a function of x, which is usually a signal telling us to try integrating by parts. This would work, but we're still going to have to integrate √(x^2+3), which is tough. Seems like integration by parts is a bit of a dead end, so we're going to have to think of something else to try. Using the substitution u = x^2+3 (when we have a square root it's often really useful to take u='whatever's in the square root') will get the right-hand side to be

= x√u dx

That's good, but we don't really want dx in there (we're integrating with respect to u now), so let's get du in terms of dx. Differentiating the expression u = x^2+3 will get us:

du/dx = 2x => dx = du/2x Substituting that into our last equation gets us:

RHS = (x√u)(du/2x) Great news! Our x's cancel, just leaving us with

the integral of √u/2 du which we can do by inspection: it's (1/2)u^3/2 divided by 3/2

Dividing by 3/2 is the same as multiplying by 2/3, so our 2's cancel and we're just left with:

e^2y/2 = 1/3(u^3/2) Oops, don't want u's in there, we're working in terms of y and x. Also we forgot to add c

e^2y/2 = 1/3(x^2+3)^3/2 + c where c is our constant of integration. That's better.

Wow; Long question! But it's a lot of marks so let's stick at it. Now to find c using our intitial values. Plugging in y = 0 when x = 1 will give us:

(e^0)/2 = 1/3(1+3)3/2 + c => 1/2 = 1/3*8 + c => c = 1/2-3/8 = 3/6-16/6 = -13/6

We're almost there. but the question wanted y = f(x) so we've got a bit of rearranging to do. Multiply by 2 first:

e^2y/2 = 1/3(x^2+3)^3/2 -13/6

=> e^2y = 2/3(x^2+3)^3/2 - 13/3 Now we're going to have to take logs to get rid of that e:

=> ln(e^2y) = ln(2/3(x^2+3)^3/2 - 13/3) ln(e^a) = a by definition, so on the left we're just left with 2y.

=> 2y = ln(2/3(x^2+3)^3/2 - 13/3) and finally we just need to divide by 2!

=> y = 1/2 ln(2/3(x^2+3)^3/2 - 13/3)

Well that question required a lot of perseverance but nothing in it was too tricky. It's so important that we can do things like integrations quickly, picking up lots of marks on questions like this and giving ourselves plenty of time for the really fiendish quesitons right at the end of the paper.

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If you're taking A-Level Maths - well done to you. It's one of the trickier A-levels but also one of the most well respected, and it will set you up for all sorts of career paths. That said, it can be a surprising jump up from the Maths you've done before. And because the course progresses so quickly, it's easy to feel like you're getting left behind.

With one-to-one online support from a Maths tutor, you can spend time focusing on the topics you want to work on.Whether it's differentiation that's got you in a pickle, or logarithms that are giving you trouble, our tutors are on hand to help. So by the time it comes round to your exams, you'll feel confident, well prepared, and able to achieve the results you want.