Given the function: $$f(θ) = 4 \, cos^2 θ - 3 \, sin^2 θ$$ (a) Show that $$f(θ) = \frac{1}{2} + \frac{7}{2} \, cos 2θ.$$ (b) Hence, using calculus, find the exact value of $$\int_{0}^{\frac{\pi}{2}} θf(θ) \, dθ.$$ - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 6

Using the result from part (a), we need to evaluate the integral:

$\int_{0}^{\frac{\pi}{2}} θf(θ) \, dθ = \int_{0}^{\frac{\pi}{2}} θ \left( \frac{1}{2} + \frac{7}{2} \, cos 2θ \right) \, dθ.$

This splits into two separate integrals:

$\int_{0}^{\frac{\pi}{2}} θ \left( \frac{1}{2} \right) \, dθ + \int_{0}^{\frac{\pi}{2}} θ \left( \frac{7}{2} \, cos 2θ \right) \, dθ.$

The first integral is:

$\int_{0}^{\frac{\pi}{2}} \frac{1}{2}θ \, dθ = \frac{1}{2} \cdot \left[ \frac{θ^2}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\frac{\pi^2}{4}}{2} = \frac{\pi^2}{16}.$

For the second integral, we can use integration by parts:

Let:

• $u = θ$
• $dv = \frac{7}{2}cos 2θ \, dθ$

Then:

• $du = dθ$
• $v = \frac{7}{4}sin 2θ$

Applying integration by parts:

$\int u \, dv = uv - \int v \, du$

Calculating: $\int_{0}^{\frac{\pi}{2}} θ \left( \frac{7}{2} cos 2θ \right) \, dθ = \left[ θ \cdot \frac{7}{4} sin 2θ \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \frac{7}{4} sin 2θ \, dθ.$

The first term evaluates to zero at both limits: $= 0 - \int_{0}^{\frac{\pi}{2}} \frac{7}{4} sin 2θ \, dθ.$

Evaluating the remaining integral: $\int sin 2θ \, dθ = -\frac{1}{2} cos 2θ.$

Thus: $\int_{0}^{\frac{\pi}{2}} sin 2θ \, dθ = -\frac{1}{2} \left[ cos 2\frac{\pi}{2} - cos 0 \right] = -\frac{1}{2}(0 - 1) = \frac{1}{2}.$

Putting it all together: $= -\frac{7}{4} \cdot \frac{1}{2} = -\frac{7}{8}.$

Final value of the integral: $\int_{0}^{\frac{\pi}{2}} θf(θ) \, dθ = \frac{\pi^2}{16} - \frac{7}{8}.$

Now converting the integral result: $= \frac{\pi^2}{16} - \frac{14}{16} = \frac{\pi^2 - 14}{16}.$