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Which expression is equal to \[\int \frac{1}{\sqrt{1-4x^2}} \, dx?\] A - HSC - SSCE Mathematics Extension 2 - Question 1 - 2018 - Paper 1

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Which-expression-is-equal-to--\[\int-\frac{1}{\sqrt{1-4x^2}}-\,-dx?\]-A-HSC-SSCE Mathematics Extension 2-Question 1-2018-Paper 1.png

Which expression is equal to \[\int \frac{1}{\sqrt{1-4x^2}} \, dx?\] A. \( \frac{1}{2} \sin^{-1}\frac{x}{2} + C \) B. \( \frac{1}{2} \sin^{-1}2x + C \) C. \( \sin^{... show full transcript

Worked Solution & Example Answer:Which expression is equal to \[\int \frac{1}{\sqrt{1-4x^2}} \, dx?\] A - HSC - SSCE Mathematics Extension 2 - Question 1 - 2018 - Paper 1

Step 1

Identify the Integral

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Answer

The integral given is ( \int \frac{1}{\sqrt{1-4x^2}} , dx ). This form resembles the integral of the inverse sine function.

Step 2

Use the Standard Integral Result

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Answer

Recall that the integral ( \int \frac{1}{\sqrt{1-u^2}} , du = \sin^{-1}(u) + C ). In this case, let ( u = 2x ), thus ( du = 2 , dx ), leading to ( dx = \frac{1}{2} du ).

Step 3

Change of Variables

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Answer

Substituting back into the integral, we have:

[ \int \frac{1}{\sqrt{1-4x^2}} , dx = \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2} , du = \frac{1}{2} \sin^{-1}(u) + C = \frac{1}{2} \sin^{-1}(2x) + C ]

Step 4

Final Answer

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Answer

Thus, the correct answer is:

B. ( \frac{1}{2} \sin^{-1}(2x) + C )

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