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

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

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

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

0

Solution steps

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

Compute the indefinite integral:

\int sin^{2} (x) cos (x) d x = \frac{sin ^{3} ( x )}{3} + C

Compute the boundaries:

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

= 0 - 0

Simplify

= 0

\int_{- 1}^{2} (3 u - 2) (u + 1) d u

\int_{0}^{1} (x^{2} - x + \frac{1}{6})^{2} d x

\int_{- 2}^{2} \frac{x + ∣ x ∣}{2 + x ^{2}} d x

\int_{0}^{2 π} x^{2} cos (x) d x

\int_{0}^{1} x^{2} (4 x^{3} - 1)^{4} d x

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