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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...

Start with a). Looking at this equation, the trig identity screaming out is cosx^2+sinx^2=1--> cosx^2=1-sinx^2. Substituting this into the LHS of the equation, and with a bit of algebraic rearrangement...