What is the integral from 0 to 1 of sin(pix)cos(pix) ?

The integral from 0 to 1 of sin(pix)cos(pix) is 0

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integral from 0 to 1 of sin(pix)cos(pix)

\int_{0}^{1} sin (π x) cos (π x) d x

0

Solution steps

\int_{0}^{1} sin (π x) cos (π x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Use the common integral:

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

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

Simplify

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

= \frac{1}{4 π} [- cos (u)]_{0}^{2 π}

Compute the boundaries:

0

= \frac{1}{4 π} \cdot 0

Simplify

= 0

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What is the integral from 0 to 1 of sin(pix)cos(pix) ?
The integral from 0 to 1 of sin(pix)cos(pix) is 0