Question

Evaluate the integral: \[ \int 4\cos^2 x - 5\sin^2 x \cos x \, dx. \]

• A. \( \frac{1}{2} \cos x \sqrt{ (4 - 9 \sin^2 x)} + \frac{2}{3} \sin^{-1} \left( \frac{3 \sin x}{2} \right) + c \)
• B. \( \frac{1}{2} \sin x \sqrt{(4 - 9 \sin^2 x)} + \frac{2}{3} \cos^{-1} \left( \frac{3 \cos x}{2} \right) + c \)
• C. \( \frac{1}{2} \cos x \sqrt{(1 - 9 \cos^2 x)} + \frac{2}{3} \sin^{-1} \left( \frac{3 \cos x}{2} \right) + c \)
• D. \( \frac{1}{2} \sin x (4 - 9 \cos^2 x) + \frac{2}{3} \sin^{-1} \left( \frac{3 \sin x}{2} \right) + c \) \bigskip

Hint:

For integrals involving powers of trigonometric functions, use substitution and trigonometric identities to simplify the problem.

Answer 1

The Correct Option is D

We start with the integral: \[ \int (4\cos^2 x - 5\sin^2 x \cos x) \, dx. \] The integral of \( \cos^2 x \) is straightforward using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \): \[ \int 4\cos^2 x \, dx = 4 \int \frac{1 + \cos 2x}{2} \, dx = 2 \int (1 + \cos 2x) \, dx. \] For \( 5 \sin^2 x \cos x \), use substitution \( u = \sin x \), then: \[ \int 5 \sin^2 x \cos x \, dx = \int 5 u^2 du = \frac{5u^3}{3}. \] Putting it all together, we have the following expression after integrating. The result is: \[ \frac{1}{2} \sin x (4 - 9 \cos^2 x) + \frac{2}{3} \sin^{-1} \left( \frac{3 \sin x}{2} \right) + c. \] \bigskip