Use the substitution u=2+sinθu = 2 + \sin \thetau=2+sinθ to show that the integral
∫24sin2θ(2+sinθ)3dθ \int \frac{24 \sin 2\theta}{(2 + \sin \theta)^3} d\theta ∫(2+sinθ)324sin2θdθcan be written in the form
∫(48u2−96u3)du \int \left( \frac{48}{u^2} - \frac{96}{u^3} \right) du ∫(u248−u396)duThe torque τ\tauτ (in N m) generated by a mechanical component is modeled by the function
τ(θ)=24sin2θ(2+sinθ)3,0≤θ≤π2 \tau(\theta) = \frac{24 \sin 2\theta}{(2 + \sin \theta)^3}, \quad 0 \le \theta \le \frac{\pi}{2} τ(θ)=(2+sinθ)324sin2θ,0≤θ≤2πwhere θ\thetaθ is the angle of rotation in radians. Calculate the exact value of the total work done, given by ∫0π/2τ(θ)dθ\int_{0}^{\pi/2} \tau(\theta) d\theta∫0π/2τ(θ)dθ.
Show each stage of your working and give your answer as a fraction in its simplest form.
363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.