Question

question 2
evaluate the indefinite integral: ∫cos⁶(x)sin(x)dx
the correct answer is not listed.
-1/7 cos⁷(x)+c
-1/7 sin⁷(x)+c
1/6 sin⁷(x)+c
1/7 sin⁷(x)+c

Explanation:

Step1: Substitute \( u = \cos(x) \)

Differentiate \( u \) with respect to \( x \): \( du=-\sin(x)dx\), so \(\sin(x)dx=-du\)
The integral \(\int\cos^{6}(x)\sin(x)dx\) becomes \(\int u^{6}(-du)=-\int u^{6}du\)

Step2: Integrate \( u^{6} \)

Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n = 6\), \(\int u^{6}du=\frac{u^{7}}{7}+C\)
So \(-\int u^{6}du=-\frac{u^{7}}{7}+C\)

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

We get \(-\frac{\cos^{7}(x)}{7}+C\)

Answer:

\(-\frac{1}{7}\cos^{7}(x)+C\)