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First, recall that when integrating, squared trigonometric functions often cause issues. Therefore, use the identity: cos^2(x) = (cos(2x) + 1)/2 to remove this power, giving the integral: ∫ (xcos(2x))/2  ...

(Question from AQA A level maths specimen papers) When looking at this question, you need to appreciate the steps required in order to reach the final answer before diving straight in. Firstly, the questi...

Step 1: multiply (5sqrt(3)-6)/(2sqrt(3)+3) by (2sqrt(3)-3)/(2sqrt(3)-3) (this is a trick called CONJUGATION as you're really just multiplying the fraction by 1. Step 2: Expand and ...

factorise the top fraction: 3x^2 - x - 2 = (3x + 2)(x - 1)

factorise the bottom fraction: x^2 -1 = (x + 1)(x - 1)

cancel out the (x - 1) and the final result is (3x + 2) / (x + 1)

to find the area integrate the function between 0 and pi/2:

indefinite integral of y = -1/2 cos(2x) + sin(x) + c

ignore c and input boundary conditions:

[-1/2 cos(pi) + sin(pi/2)] - [...