What is the integral of sin(mpix) ?

The integral of sin(mpix) is -1/(pim)cos(pimx)+C

Popular >

integral of sin(mpix)

\int sin (mπ x) d x

- \frac{1}{πm} cos (πm x) + C

Solution steps

\int sin (mπ x) d x

Apply u-substitution

: \int \frac{sin ( u )}{πm} d u

= \int \frac{sin ( u )}{πm} d u

Take the constant out:

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

= \frac{1}{πm} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{πm} (- cos (u))

Substitute back

u = m x π

= \frac{1}{πm} (- cos (m x π))

Simplify

= - \frac{1}{πm} cos (πm x)

Add a constant to the solution

= - \frac{1}{πm} cos (πm x) + C

f (x) = 2 x - \frac{e ^{x}}{2}

\int_{π}^{\infty} \frac{4}{x ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + y = (3 x + 4) e^{3 x}

f (x) = x^{5} - x^{3} + 3

y = x^{4} + 9 e^{x}, (0, 9)

What is the integral of sin(mpix) ?
The integral of sin(mpix) is -1/(pim)cos(pimx)+C