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

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

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

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

0

Solution steps

\int_{0}^{2 π} sin (x) cos (x) d x

Rewrite using trig identities

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

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 x) d x

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{π}{4}}^{0} \frac{1}{cos ^{2} ( x )} d x

\int_{0}^{y} \frac{1}{2 y} d y

\int_{0}^{2} 3 x^{2} + 1 d x

\int_{s i n (x)}^{c o s (x)} x^{2} d x

\int_{0}^{\frac{π}{2}} x cos (x) d x

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