What is the integral of sin(pix)cos(pix) ?

The integral of sin(pix)cos(pix) is -1/(4pi)cos(2pix)+C

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integral of sin(pix)cos(pix)

\int sin (π x) cos (π x) d x

- \frac{1}{4 π} cos (2 π x) + C

Solution steps

\int sin (π x) cos (π x) d x

Rewrite using trig identities

= \int \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 sin (2 π x) d x

Apply u-substitution

: \int sin (u) \frac{1}{2 π} d u

= \frac{1}{2} \cdot \int 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 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))

Substitute back

u = 2 π x

= \frac{1}{2} \cdot \frac{1}{2 π} (- cos (2 π x))

Simplify

\frac{1}{2} \cdot \frac{1}{2 π} (- cos (2 π x)) : - \frac{1}{4 π} cos (2 π x)

= - \frac{1}{4 π} cos (2 π x)

Add a constant to the solution

= - \frac{1}{4 π} cos (2 π x) + C

\frac{d}{d x} (cosh (a x))

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

\frac{\partial}{\partial x} (\frac{2 \cdot x}{x ^{2} - y ^{2}})

What is the integral of sin(pix)cos(pix) ?
The integral of sin(pix)cos(pix) is -1/(4pi)cos(2pix)+C