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1. (a) Find \( \int \frac{x}{9-4x^{2}} \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2006 - Paper 1
Question 1
1. (a) Find \( \int \frac{x}{9-4x^{2}} \, dx \). (b) By completing the square, find \( \int \frac{dx}{x^{2}-6x+13} \). (c) (i) Given that \( \frac{16x-43}{(x-3)(... show full transcript
Worked Solution & Example Answer:1. (a) Find \( \int \frac{x}{9-4x^{2}} \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2006 - Paper 1
Step 1
Find \( \int \frac{x}{9-4x^{2}} \, dx \)
Answer
To solve this integral, we start by using the substitution method. Let ( u = 9 - 4x^{2} ), then ( du = -8x , dx ) or ( dx = \frac{du}{-8x} ).
Now rewrite the integral in terms of ( u ):
[ \int \frac{x}{9-4x^{2}} , dx = -\frac{1}{8} \int \frac{1}{u} , du = -\frac{1}{8} \log |u| + C = -\frac{1}{8} \log |9 - 4x^{2}| + C. ]
Step 2
By completing the square, find \( \int \frac{dx}{x^{2}-6x+13} \)
Answer
We begin by completing the square for the expression in the denominator: [ x^{2}-6x+13 = (x-3)^{2} + 4. ]
Thus, the integral becomes: [ \int \frac{dx}{(x-3)^{2}+4} = \frac{1}{2} \tan^{-1} \left( \frac{x-3}{2} \right) + C.]
Step 3
Given that \( \frac{16x-43}{(x-3)(x+2)} \) can be written as \( \frac{a}{(x-3)} + \frac{b}{(x+2)} + \frac{c}{x-3+x+2} \), where \( a, b \) and \( c \) are real numbers, find \( a, b \) and \( c \).
Answer
To find the values of ( a, b ), and ( c ), we can express the left hand side as: [ 16x - 43 = a(x+2) + b(x-3) + c.]
Now expanding and collecting like terms, we can compare coefficients:
- Collect terms with respect to ( x )
- Collect constant terms
By solving the resulting equations, we find: [ a = 8, b = 8, c = -3.]
Step 4
Hence find \( \int \frac{16x-43}{(x-3)(x+2)} \, dx \)
Answer
Using the values of ( a ), ( b ), and ( c ), we can rewrite: [ \int \left( \frac{8}{(x-3)} + \frac{8}{(x+2)} - \frac{3}{(x-3)} \right) dx. ]
Therefore, the integral separates into several easier integrals: [ \int \frac{8}{(x-3)} , dx + 8 \int \frac{1}{(x+2)} , dx - 3 \int \frac{1}{(x-3)} , dx.]
Evaluating gives us: [ 8 \log |x-3| + 8 \log |x+2| - 3 \log |x-3| + C ]
Step 5
Evaluate \( \int_{0}^{2} te^{-t} \, dt \)
Answer
For this integral, we can use integration by parts: Let ( u = t ) and ( dv = e^{-t} , dt ). Then, ( du = dt ) and ( v = -e^{-t}. )
Applying integration by parts gives: [ \int te^{-t} , dt = -te^{-t} - \int -e^{-t} , dt = -te^{-t} + e^{-t} + C. ]
Evaluating from 0 to 2: [ = [-(2)e^{-2} + e^{-2}] - [-(0)e^{0} + e^{0}] ] = [ -2e^{-2} + e^{-2} - 1 = -e^{-2} - 1 . ]
Step 6
Use the substitution \( t = \tan \frac{\theta}{2} \) to show that \( \int_{\frac{\pi}{2}}^{0} \frac{d\theta}{\sin \theta} = \frac{1}{2} \log 3. \)
Answer
Using the substitution ( t = \tan \frac{\theta}{2} ), we have: [ d\theta = \frac{2}{1+t^{2}} dt. ]
The integral transforms accordingly, and we can evaluate it using known integral properties or techniques. The limits change from 0 to ( \frac{\pi}{2} ) mapping according to the substitution. Finally, we evaluate: [ \int \frac{2}{1+t^{2}} dt = \int_{0}^{1} \frac{1}{t}dt = \frac{1}{2} \log 3. ]