Question

Evaluate the integral: \( \int \sin^5 x \, dx \)

• A. \( -\frac{1}{5} \cos x (5 - 10 \sin^2 x + \sin^4 x) + C \)
• B. \( -\cos x + \frac{\cos^3 x}{3} - \frac{\cos^5 x}{5} + C \)
• C. \( \frac{1}{5} \sin^5 x + C \)
• D. \( \int \sin^5 x \, dx = \int \sin^3 x \cdot \sin^2 x \, dx \)

Hint:

When integrating odd powers of sine or cosine, save one sine (or cosine) factor and convert the rest using trigonometric identities.

Answer 1

The Correct Option is A

\( \int \sin^5 x \, dx \), we break it into: \[ \int \sin^5 x \, dx = \int \sin^3 x \cdot \sin^2 x \, dx = \int \sin^3 x \cdot (1 - \cos^2 x) \, dx \] Now, use \( \sin^3 x = \sin x (1 - \cos^2 x) \): \[ = \int \sin x (1 - \cos^2 x)^2 (1 - \cos^2 x) \, dx = \int \sin x (1 - 2 \cos^2 x + \cos^4 x) \, dx \] Now let \( u = \cos x \), \( du = -\sin x \, dx \), so the integral becomes: \[ = -\int (1 - 2u^2 + u^4) \, du = -\left( u - \frac{2u^3}{3} + \frac{u^5}{5} \right) + C \] \[ = -\cos x + \frac{2}{3} \cos^3 x - \frac{1}{5} \cos^5 x + C \]