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

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

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

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

0

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{0}^{2 π} sin (x) cos^{2} (x) d x = - \frac{1}{3} - (- \frac{1}{3})

= - \frac{1}{3} - (- \frac{1}{3})

Simplify

= 0

\int sec (y) (tan (y) - sec (y)) d y

\frac{d}{d x} (e^{(x + 1)^{2}})

x \to π lim (ln (cos (x) - 5))

f (x) = x^{2} + 3, at x = \frac{1}{2}

\frac{\partial}{\partial y} (cos (x z))

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