Consider the function: $$f(\theta) = 4\cos^2\theta - 3\sin^2\theta$$ (a) Show that $$f(\theta) = \frac{1}{2} - \frac{7}{2}\cos 2\theta.$$ (b) Hence, using calculus, find the exact value of $$\int_0^{\frac{\pi}{2}} \theta f(\theta) d\theta.$$ - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 6

To solve the integral, we first plug in $f(\theta)$:

$\int_0^{\frac{\pi}{2}} \theta f(\theta) d\theta = \int_0^{\frac{\pi}{2}} \theta \left(\frac{1}{2} - \frac{7}{2}\cos 2\theta\right) d\theta$
This can be split into two integrals:

$= \frac{1}{2}\int_0^{\frac{\pi}{2}} \theta d\theta - \frac{7}{2}\int_0^{\frac{\pi}{2}} \theta \cos 2\theta d\theta$
For the first integral:

$\int_0^{\frac{\pi}{2}} \theta d\theta = \left[\frac{\theta^2}{2}\right]_0^{\frac{\pi}{2}} = \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{\pi^2}{8}$

Now for the second integral, we can use integration by parts: Let:

• $u = \theta$
• $dv = \cos 2\theta d\theta$
  Then:
• $du = d\theta$
• $v = \frac{1}{2}\sin 2\theta$
  By applying integration by parts:

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

Thus:

$\int \theta \cos 2\theta d\theta = \left[\frac{\theta}{2}\sin 2\theta\right]_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \frac{1}{2}\sin 2\theta d\theta$
The boundary term evaluates to 0:

$\left[\frac{\theta}{2}\sin 2\theta\right]_0^{\frac{\pi}{2}} = 0$
Next, we need to evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{2}\sin 2\theta d\theta$:

This gives: $\int \sin 2\theta d\theta = -\frac{1}{2}\cos 2\theta$
So: $\int_0^{\frac{\pi}{2}} \frac{1}{2}\sin 2\theta d\theta = \left[-\frac{1}{4}\cos 2\theta\right]_0^{\frac{\pi}{2}} = -\frac{1}{4}(\cos\pi - \cos 0) = -\frac{1}{4}(-1 - 1) = \frac{1}{2}$
Thus:

$\int \theta \cos 2\theta d\theta = 0 - \frac{1}{2} = -\frac{1}{2}$
Putting everything back together yields:

$\int_0^{\frac{\pi}{2}} \theta f(\theta) d\theta = \frac{1}{2}\cdot\frac{\pi^2}{8} - \frac{7}{2}\cdot(-\frac{1}{2})$
This simplifies to:

$= \frac{\pi^2}{16} + \frac{7}{4} = \frac{\pi^2 + 28}{16}$
Therefore, the exact value is:

$\int_0^{\frac{\pi}{2}} \theta f(\theta) d\theta = \frac{\pi^2 + 28}{16}.$