Find $$ \int \frac{\cos \theta}{\sin^2 \theta} \, d\theta $$ (ii) Find real numbers \( a \) and \( b \) such that $$ \frac{5x}{x^2 - x - 6} = \frac{a}{x - 3} + \frac{b}{x + 2} $$ (iii) 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} $$ (i) Show that $$ \frac{dt}{d\theta} = \frac{1}{2}(1 + t^2) $$ (ii) Show that $$ \sin \theta = \frac{2t}{1 + t^2} $$ (iii) Use the substitution \( t = \tan \frac{\theta}{2} $$ to find $$ \int \csc \theta \, d\theta $$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2005 - Paper 1

To find ( a ) and ( b ), we equate coefficients:

1. ( a + b = 5 )
2. ( -3a + 2b = -1 )

Solving these equations, we get ( a = 2 ) and ( b = 3 ).

Using the values of ( a ) and ( b ) found earlier, we have:

$\int \left( \frac{2}{x - 3} + \frac{3}{x + 2} \right) \, dx = 2 \ln |x - 3| + 3 \ln |x + 2| + C$. This simplifies to: $\ln |(x - 3)^2 (x + 2)^3| + C$

Let ( u = log_e x ) and ( dv = x^7 , dx ). Then, ( du = \frac{1}{x} , dx ) and ( v = \frac{x^8}{8} ). Using integration by parts:

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

Substituting the values, we get:

$\left[ log_e x \cdot \frac{x^8}{8} \right]_1^e - \int \frac{x^8}{8} \cdot \frac{1}{x} \, dx$. Evaluating gives: $\frac{1}{8} (e^8 \cdot 1 - 0) - \frac{1}{8} \int x^7 \, dx = \frac{1}{8} (e^8 - 0).$

This integral can be solved using a standard integral formula:

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \left( \frac{x}{a} \right) + C$

Here, setting ( a = 2 ), we can write: $\int \frac{dx}{\sqrt{4x^2 - 1}} = \frac{1}{2} \ln \left| 2x + \sqrt{4x^2 - 1} \right| + C$.

Using the chain rule, differentiate ( t = \tan \frac{\theta}{2} ):

$\frac{dt}{d\theta} = \frac{1}{2} sec^2 \frac{\theta}{2}$

Since ( sec^2 x = 1 + tan^2 x ), we have:

$\frac{dt}{d\theta} = \frac{1}{2} (1 + tan^2 \frac{\theta}{2}) = \frac{1}{2} (1 + t^2).$

Using the double angle formula for sine, we have:

$\sin \theta = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} = \frac{2t}{1 + t^2}.$

Noting that:

$\csc \theta = \frac{1}{\sin \theta} = \frac{1 + t^2}{2t}$

Using the substitution:

$d\theta = \frac{2}{1 + t^2} dt$

Thus:

$\int \csc \theta \, d\theta = \int \frac{(1 + t^2)}{2t} \cdot \frac{2}{1 + t^2} dt = \int \frac{1}{t} dt = \ln |t| + C = \ln |tan \frac{\theta}{2}| + C$.