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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

Question 2

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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. \[ \int \frac{1 - \cos 6x}{2} \, dx \] ... 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 using the identity

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Answer

To find an equivalent integral, we can apply the identity for sine squared:

[ \sin^2 A = \frac{1 - \cos(2A)}{2} ]

In this case, we have:

[ A = 3x \implies \sin^2 3x = \frac{1 - \cos(6x)}{2} ]

Thus, substituting into the integral:

[ \int \sin^2 3x , dx = \int \frac{1 - \cos 6x}{2} , dx ]

Step 2

Identify the correct option

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Answer

From the options provided, we find:

B. [ \int \frac{1 - \cos 6x}{2} , dx ]

is equivalent to ( \int \sin^2 3x , dx ). Therefore, the correct answer is B.

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