Question

question 20
evaluate the indefinite integral.
$\int\cos ^{5}x\sin xdx$
$\frac{1}{6}\cot ^{6}x + c$
$-\frac{1}{6}\sin ^{6}x + c$
$-\frac{1}{6}\cos ^{6}x + c$
$\frac{1}{6}\sin ^{6}x + c$
$\frac{1}{6}\cos ^{6}x + c$

Explanation:

Step1: Use substitution

Let \(u = \cos x\), then \(du=-\sin xdx\), and \(\sin xdx=-du\). The integral \(\int\cos^{5}x\sin xdx\) becomes \(-\int u^{5}du\).

Step2: Integrate \(u^{5}\)

Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 5\), we have \(-\int u^{5}du=-\frac{u^{6}}{6}+C\).

Step3: Substitute back \(u=\cos x\)

Substituting \(u = \cos x\) back into the result, we get \(-\frac{\cos^{6}x}{6}+C\).

Answer:

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