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

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

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integral from 0 to 2pi of cos(t)sin(t)

\int_{0}^{2 π} cos (t) sin (t) d t

0

Solution steps

\int_{0}^{2 π} cos (t) sin (t) d t

Rewrite using trig identities

= \int_{0}^{2 π} \frac{sin ( 2 t )}{2} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{2 π} sin (2 t) d t

Apply u-substitution

: \int_{0}^{4 π} sin (u) \frac{1}{2} d u

= \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) \frac{1}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{4 π} sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π}

Simplify

\frac{1}{2} \cdot \frac{1}{2} [- cos (u)]_{0}^{4 π} : \frac{1}{4} [- cos (u)]_{0}^{4 π}

= \frac{1}{4} [- cos (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{1}{4} \cdot 0

Simplify

= 0

\int \frac{3}{2} (x + 1)^{\frac{1}{2}} d x

x \to - \frac{1}{2} lim (\frac{( 6 x - 1 )}{x})

1 7^{x}

\frac{d}{d x} (π - x)

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