1. Find $$\int \frac{\cos \theta}{\sin^2 \theta} d\theta.$$ (b) (i) Find real numbers $a$ and $b$ such that $$\frac{5x}{x^2 - x - 6} = \frac{a}{x - 3} + \frac{b}{x + 2}.$$ (ii) Hence find $$\int \frac{5x}{x^2 - x - 6} dx.$$ (c) Use integration by parts to evaluate $$\int_1^e x^7 \log_e x dx.$$ d) Using the table of standard integrals, or otherwise, find $$\int \frac{dx}{\sqrt{4x^2 - 1}}.$$ e) Let $t = \tan \frac{\theta}{2}$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2005 - Paper 1

To find $a$ and $b$, we can equate coefficients.
Thus:
$\frac{5x}{(x - 3)(x + 2)} = \frac{a(x + 2) + b(x - 3)}{(x - 3)(x + 2)}.$
This leads to:
$5x = a(x + 2) + b(x - 3).$
Expanding the right side gives:
$5x = (a + b)x + (2a - 3b).$
Setting coefficients equal gives two equations, $a + b = 5$ and $2a - 3b = 0$. Solving these we find $a = 2$ and $b = 3$.

Using the previously found values of $a$ and $b$, we rewrite our integral:
$\int \left( \frac{2}{x - 3} + \frac{3}{x + 2} \right) dx = 2 \ln |x - 3| + 3 \ln |x + 2| + C.$

Using integration by parts, let $u = \log_e x$ and $dv = x^7 dx$.
Then, $du = \frac{1}{x} dx$ and $v = \frac{x^8}{8}$.
Using the formula $\int u \, dv = uv - \int v \, du$, we have:
$\int_1^e x^7 \log_e x dx = \left[ \frac{x^8}{8} \log_e x \right]_1^e - \int_1^e \frac{x^8}{8} \cdot \frac{1}{x} dx.$
Evaluating these gives the result, simplifying gives:
$= e\cdot \frac{e^8}{8} - \frac{1}{8} \int_1^e x^7 dx.$
The remaining integral can be computed separately, leading to the final evaluation.

Recognizing this as a standard integral, we can rewrite it as:
$\int \frac{dx}{\sqrt{4x^2 - 1}} = \frac{1}{2} \ln \left| 2x + \sqrt{4x^2 - 1} \right| + C.$

Starting with the given substitution $t = \tan \frac{\theta}{2}$:
Using the derivative of $\tan$, we apply the chain rule:
$\frac{dt}{d\theta} = \frac{1}{2} \sec^2 \frac{\theta}{2} = \frac{1}{2} \left( 1 + \tan^2 \frac{\theta}{2} \right) = \frac{1}{2} (1 + t^2).$

Using the half-angle identities, we can derive this as follows:
$\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} = \frac{2t}{1 + t^2}.$

Substituting the expression for $\sin \theta$ gives us:
$\int \csc \theta d\theta = \int \frac{1}{\sin \theta} d\theta = \int \frac{1 + t^2}{2t} d\theta.$
This will require converting in terms of $t$, leading to the final integral evaluation.