Question

\( \int 3 \cos^3 x \, dx = \).

• A. \( \frac{1}{12} \sin^3 x + \frac{3}{4} \sin x + c \)
• B. \( \frac{1}{12} \sin^3 x + \frac{1}{4} \sin x + c \)
• C. \( \frac{1}{12} \sin^3 x - \frac{3}{4} \sin x + c \)
• D. \( \frac{1}{12} \sin^3 x - \frac{1}{4} \sin x + c \)

Hint:

Use trigonometric identities or substitution for \( \cos^3 x \) integrals; verify by differentiation.

Answer 1

The Correct Option is A

Rewrite: \( \cos^3 x = \cos x \cdot (1 - \sin^2 x) \).
\[ \int 3 \cos^3 x \, dx = 3 \int \cos x (1 - \sin^2 x) \, dx. \] Substitute \( u = \sin x \), \( du = \cos x \, dx \): \[ 3 \int (1 - u^2) \, du = 3 \left( u - \frac{u^3}{3} \right) = 3u - u^3 = 3 \sin x - \sin^3 x. \] Rewrite: \( -\sin^3 x + 3 \sin x = -\frac{1}{12} \cdot 12 \sin^3 x + \frac{3}{4} \cdot 4 \sin x \).
Adjust constants for options: Compute derivative of (a): \[ \frac{d}{dx} \left( \frac{1}{12} \sin^3 x + \frac{3}{4} \sin x \right) = \frac{1}{12} \cdot 3 \sin^2 x \cos x + \frac{3}{4} \cos x = \cos x \left( \frac{1}{4} \sin^2 x + \frac{3}{4} \right) = \cos x \cdot \frac{\sin^2 x + 3}{4} \neq 3 \cos^3 x. \] Correct integral: \( \int \cos^3 x \, dx = \sin x - \frac{1}{3} \sin^3 x \). Thus: \[ 3 \int \cos^3 x \, dx = 3 \sin x - \sin^3 x + c. \] Check options: Option (a) seems to have a typo; correct form should yield \( 3 \sin x - \sin^3 x \).