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Which of the following integrals is equivalent to $$\int \sin^{2}3x\,dx?$$ A - HSC - SSCE Mathematics Extension 1 - Question 2 - 2021 - Paper 1

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Which of the following integrals is equivalent to $$\int \sin^{2}3x\,dx?$$ A. \int \frac{1 + \cos 6x}{2}\,dx B. \int \frac{1 - \cos 6x}{2}\,dx C. \int \frac{1 +... show full transcript

Worked Solution & Example Answer:Which of the following integrals is equivalent to $$\int \sin^{2}3x\,dx?$$ A - HSC - SSCE Mathematics Extension 1 - Question 2 - 2021 - Paper 1

Step 1

Evaluate the Integral

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Answer

To solve the integral ∫sin⁡23x dx\int \sin^{2}3x\,dx, we can use the identity: sin⁡2θ=1−cos⁡(2θ)2\sin^{2}\theta = \frac{1 - \cos(2\theta)}{2}. Substituting (\theta = 3x), we have: sin⁡23x=1−cos⁡(6x)2\sin^{2}3x = \frac{1 - \cos(6x)}{2}. Thus, we rewrite the integral as: ∫sin⁡23x dx=∫1−cos⁡(6x)2 dx.\int \sin^{2}3x\,dx = \int \frac{1 - \cos(6x)}{2}\,dx.

Step 2

Identify the Equivalent Integral

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Answer

From the evaluation above, we see that: ∫1−cos⁡(6x)2 dx\int \frac{1 - \cos(6x)}{2}\,dx equals option B: B. \int \frac{1 - \cos 6x}{2},dx.

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