What is the integral from 0 to 2pi of cos^5(x)sin(x) ?

The integral from 0 to 2pi of cos^5(x)sin(x) is 0

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integral from 0 to 2pi of cos^5(x)sin(x)

\int_{0}^{2 π} cos^{5} (x) sin (x) d x

0

Solution steps

\int_{0}^{2 π} cos^{5} (x) sin (x) d x

Compute the indefinite integral:

\int cos^{5} (x) sin (x) d x = - \frac{cos ^{6} ( x )}{6} + C

Compute the boundaries:

\int_{0}^{2 π} cos^{5} (x) sin (x) d x = - \frac{1}{6} - (- \frac{1}{6})

= - \frac{1}{6} - (- \frac{1}{6})

Simplify

= 0

\int_{0}^{1} x^{2} (ln (\frac{1}{x}))^{3} d x

\int_{0}^{2} 4 - x d x

\int_{0}^{2} - g \cdot cos (n \cdot π \cdot g) d g

\int_{- 1}^{2} y^{2} + 4 d y

\int_{0}^{2} 20 x d x

What is the integral from 0 to 2pi of cos^5(x)sin(x) ?
The integral from 0 to 2pi of cos^5(x)sin(x) is 0