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Question 1 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 1 - 2008 - Paper 1
Question 1
Question 1 (15 marks) Use a SEPARATE writing booklet. (a) Find \( \int \frac{x^2}{(5+x^2)^2} \: dx \). (b) Find \( \int \frac{dx}{\sqrt{4x^2+1}} \). (c) Evaluate ... show full transcript
Worked Solution & Example Answer:Question 1 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 1 - 2008 - Paper 1
Step 1
Find \( \int \frac{x^2}{(5+x^2)^2} \: dx \)
Answer
To solve this integral, we can use the substitution method. Let ( u = 5 + x^2 ). Then, ( du = 2x , dx ) or ( dx = \frac{du}{2x} ), and we replace ( x^2 ) with ( u - 5 ).
This gives us:
[ \int \frac{x^2}{(5+x^2)^2} : dx = \int \frac{u-5}{u^2} \cdot \frac{du}{2x} = \frac{1}{2} \int \frac{u-5}{u^2} du]
Now we can separate this integral and evaluate:
[ \frac{1}{2} \left( \int \frac{1}{u} du - 5 \int \frac{1}{u^2} du \right) = \frac{1}{2} (\ln |u| + \frac{5}{u}) + C]
Substituting back ( u = 5 + x^2 ) results in:
[ \frac{1}{2} \left( \ln(5+x^2) + \frac{5}{5+x^2} \right) + C ]
Step 2
Step 3
Evaluate \( \int_0^1 \tan^{-1} x \, dx \)
Answer
To evaluate this integral, we can use integration by parts. Let:
[ u = \tan^{-1} x, , dv = dx \Rightarrow du = \frac{1}{1+x^2} dx , and : v = x ]
Now applying integration by parts:
[ \int u , dv = uv - \int v , du ]
Evaluate at limits 0 and 1 gives:
[ \left[ x \tan^{-1} x \right]_0^1 - \int_0^1 \frac{x}{1+x^2} dx = \frac{\pi}{4} - \frac{1}{2} \ln(2) ]
Step 4
Evaluate \( \int_2^0 \frac{dx}{\sqrt{2x-1}} \)
Answer
For this integral, we can switch the limits and evaluate:
[ \int_2^0 \frac{dx}{\sqrt{2x-1}} = - \int_0^2 \frac{dx}{\sqrt{2x-1}}]
Using the substitution ( x = \frac{1}{2}u^2 + \frac{1}{2} ), we can evaluate this integral to yield the final form:
[ - \ln(\sqrt{2x-1}+1) \text{ evaluated between 0 and 2} ]
Step 5
Use this result to evaluate \( \int_0^1 \frac{8(1-x)}{(2-x^2)(2-2x+x^2)} \: dx \)
Answer
Using the given result and simplifying, we can write the integral as:
[ \int_0^1 \frac{8(1-x)}{(2-x^2)(2-2x+x^2)} : dx = \int_0^1 \left( \frac{4-2x}{2-2x+x^2} - \frac{2x}{2-x^2} \right) dx]
Evaluating each term separately will give us the final answer.