Question

evaluate the indefinite integral: $int cos^{5} x sin x , dx$.

$-\frac{1}{6}sin^{6} x + c$

$\frac{1}{6}cot^{6} x + c$

$\frac{1}{6}cos^{6} x + c$

$\frac{1}{6}sin^{6} x + c$

$-\frac{1}{6}cos^{6} x + c$

Explanation:

Step1: Choose Substitution

Let \( u = \cos x \), then \( du = -\sin x \, dx \), so \( -\!du = \sin x \, dx \).

Step2: Substitute into Integral

The integral becomes \( \int u^5 (-\!du) = -\int u^5 du \).

Step3: Integrate

Integrate \( u^5 \): \( -\frac{u^6}{6} + C \).

Step4: Substitute Back

Replace \( u \) with \( \cos x \): \( -\frac{1}{6}\cos^6 x + C \).

Answer:

\( -\frac{1}{6}\cos^6 x + C \) (the last option: \( -\frac{1}{6}\cos^6 x + C \))