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We can break this down into steps, going deeper each time. First we might just say: well, since integration is fundamentally the inverse process of differentiation and we know that the derivative of ln(x)...

Stationary points are points at which the gradient of the curve is zero. The gradient is given by dy/dx so we start by computing this using product rule to give us dy/dx=-sin²x+cos²x...

There are two stages to this integral. Stage 1: Notice that there is a quadratic inside a square root in the denominator. We wish to look for a substitution which will reduce this problem...

Step One: stationary/turning points are points on the curve where the gradient equals 0 (i.e. a point at which the slope changes from negative to positive, or vice versa). So we ne...

a) Using log rules, a^x=b becomes log(a)b=x. If we take ln of both sides, we get ln(e^(3x-9))=ln8. ln(e) =1, so we just get 3x-9=ln8. Now we can simply manipulate this to get x=(9+ln8)/3=3+(1/3)ln8. Anoth...