The indefinite integral ∫(cosx3−3sin3x)dx[math] \int(\cos \frac{x}{3} -3\sin3x)dx is equal to:

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We start by restating the integral in question to confirm we are integrating the same expression: ∫(cos(x/3) − 3 sin(3x)) dx.

Option a: -(3) sin(x/3) + cos(3x) + c. If we differentiate this, we get -3 * cos(x/3) * (1/3) + (-sin(3x)) * 3 by chain rule, which simplifies to -cos(x/3) − 3 sin(3x). This does not match the original integrand cos(x/3) − 3 sin(3x), so option a is incorrect.

Option b: sin(x/3) + cos(3x) + c. Differentiating gives (1/3) cos(x/3) − 3 sin(3x). The first term is (1/3) cos(x/3), which does not equal cos(x/3), so this option is incorrect.

Option c: 3 sin(x/3) + cos(3x) + c. Differentiating yields 3 * (1/3) cos(x/3) − 3 * sin(3x) * 3 by chain rule. This simplifies to cos(x/3) − 9 sin(3x). This would give -9 sin(3x) instead of -3 sin(3x), so as written it seems off at first glance; however, carefully applying the derivative of cos(3x) is -sin(3x) * 3, which is -3 sin(3x). The full derivative is cos(x/3) − 3 sin(3x), matching the original integrand. Therefore option c is consistent with the integrand.

Option d: 3 sin(x/3) − cos(3x) + c. Differentiating gives (3) * (1/3) cos(x/3) − (-sin(3x)) * 3, which simplifies to cos(x/3) + 9 sin(3x). This differs from the original integrand due to the sign and coefficient on the sin(3x) term, so option d is incorrect.

In summary, after checking each option against the derivative being the original integrand, option c accurately reproduces cos(x/3) − 3 sin(3x) upon differentiation, while the other options fail due to incorrect signs or coefficients in the differentiated terms.

MUF0092 Mathematics Unit 2 - Semester 2, 2025 3.8 Study Area 1 Revision Quiz

The indefinite integral ∫(cosx3−3sin3x)dx[math] \int(\cos \frac{x}{3} -3\sin3x)dx is equal to:

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