QCAA Specialist Mathematics Integration techniques

$\frac{22}{(2x-3)(x+4)} = \frac{A}{2x-3} + \frac{B}{x+4}$
$\therefore 22 = A(x+4) + B(2x-3)$
Let $x = \frac{3}{2}: A(\frac{3}{2}+4) = 22 \Rightarrow A = 4$
Let $x = -4: B(2 \times -4 - 3) = 22 \Rightarrow B = -2$
$\int \frac{22}{2x^2+5x-12} dx = \int \frac{4}{2x-3} dx + \int \frac{-2}{x+4} dx$
$= 2 \int \frac{2}{2x-3} dx - 2 \int \frac{1}{x+4} dx$
$= 2 \ln|2x-3| - 2 \ln|x+4| + c$

$\int_{-3}^{0} \frac{22}{(2x-3)(x+4)} dx$
$= [2 \ln|2x-3| - 2 \ln|x+4|]_{-3}^{0}$
$= ((2 \ln|-3| - 2 \ln|4|) - (2 \ln|-9| - 2 \ln|1|))$
$= 2(\ln(3) - \ln(4) - \ln(9)) = 2 \ln(\frac{3}{9 \times 4})$
$= 2 \ln(\frac{1}{12})$

$\frac{dv}{dt} = -(4+v^2)$

$\frac{1}{4+v^2} \frac{dv}{dt} = -1$

$\int \frac{1}{4+v^2} dv = \int -1 dt$

$\frac{1}{2} \int \frac{2}{4+v^2} dv = \int -1 dt$

$\frac{1}{2} \tan^{-1}\left(\frac{v}{2}\right) = -t + c$

Given $v(0) = 1.5$
$\frac{1}{2} \tan^{-1}\left(\frac{1.5}{2}\right) = c$
$c \approx 0.32$

When $v=0$:
$\frac{1}{2} \tan^{-1}(0) \approx -t + 0.32$
$t \approx 0.32 \text{ s}$
The particle comes to rest after 0.32 s.

$\int_{-a}^a 1 + \left(\frac{\sec(2x) + \tan(2x)}{\text{cosec}(2x) + 1}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\frac{1}{\cos(2x)} + \frac{\sin(2x)}{\cos(2x)}}{\frac{1}{\sin(2x)} + 1}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\frac{1 + \sin(2x)}{\cos(2x)}}{\frac{1 + \sin(2x)}{\sin(2x)}}\right)^2 dx = 1$
$\int_{-a}^a 1 + \left(\frac{\sin(2x)}{\cos(2x)}\right)^2 dx = 1$
$\int_{-a}^a 1 + (\tan(2x))^2 dx = 1$
$\int_{-a}^a \sec^2(2x)dx = 1$
$\frac{1}{2}\tan(2x)\Big|_{-a}^a = 1$
$\frac{1}{2}(\tan(2a) - \tan(-2a)) = 1$
$\frac{1}{2}(\tan(2a) + \tan(2a)) = 1$
$\frac{1}{2}(2\tan(2a)) = 1 \Rightarrow 2a = \tan^{-1}(1)$
$2a = \frac{\pi}{4} \Rightarrow a = \frac{\pi}{8}$