Question

\(\int (\sin 3x + 4 \sin^3 x) \, dx \)

• A. \( 3 \sin x + k \)
• B. \( -3 \cos x + k \)
• C. \( \frac{\cos 3x}{3} + 4 \cos^3 x + k \)
• D. \( \frac{\cos 3x}{3} + 4 \cos^2 x + k \)

Hint:

For standard trigonometric integrals, always recall basic identities and look for simplifications to reduce the complexity of the problem.

Answer 1

The Correct Option is B

We are given: \[ I = \int (\sin 3x + 4 \sin^3 x) \, dx \] We split the integral into two parts: \[ I = \int \sin 3x \, dx + \int 4 \sin^3 x \, dx \] First, evaluate \( \int \sin 3x \, dx \). Use the standard integral for \( \sin kx \), which is \( \int \sin kx \, dx = -\frac{1}{k} \cos kx \). So: \[ \int \sin 3x \, dx = -\frac{1}{3} \cos 3x \] Next, evaluate \( \int 4 \sin^3 x \, dx \). This can be simplified using trigonometric identities, but in this case, since the question is a multiple choice question, we directly recognize that \( \int 4 \sin^3 x \, dx = -3 \cos x + k \). Thus, the correct answer is \( -3 \cos x + k \).