What is the integral of 1/pi sin(pix) ?

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

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integral of 1/pi sin(pix)

\int \frac{1}{π} sin (π x) d x

- \frac{1}{π ^{2}} cos (π x) + C

Solution steps

\int \frac{1}{π} sin (π x) d x

Take the constant out:

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

= \frac{1}{π} \cdot \int sin (π x) d x

Apply u-substitution

: \int sin (u) \frac{1}{π} d u

= \frac{1}{π} \cdot \int sin (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \frac{1}{π} \cdot \int sin (u) d u

Use the common integral:

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

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

Substitute back

u = π x

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

Simplify

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

= - \frac{1}{π ^{2}} cos (π x)

Add a constant to the solution

= - \frac{1}{π ^{2}} cos (π x) + C

y = 4 x

y = \frac{x}{8 + x}

\frac{d}{d x} (cot^{7} (x))

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 4

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

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