WAEC 2020 · Paper 2 · Q11

(a)
Find the positive value of $y$ for which $\begin{vmatrix} y & 1 & 0 \\ 1 & y & 1 \\ 0 & 1 & y \end{vmatrix} = 0$ (4 d.p.).
(b)
Evaluate $\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{4\cos\theta}{3 + 2\sin\theta}\,d\theta$ (4 d.p.).

Worked solution (try it first)

Let $u = 3 + 2\sin\theta$ .
Then $du = 2\cos\theta\,d\theta$ , so $4\cos\theta\,d\theta = 2\,du$ .
Change the limits: $\theta = -\frac\pi2$ gives $u = 1$ , and $\theta = \frac\pi2$ gives $u = 5$ .
$\displaystyle\int_1^5 \frac{2}{u}\,du = [2\ln u]_1^5$
$= 2\ln5 - 2\ln1$
$= 2\ln5$
$\approx 3.2189$ .