What is the integral of sin((npix)/1) ?

The integral of sin((npix)/1) is -1/(pin)cos(pinx)+C

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integral of sin((npix)/1)

\int sin (\frac{nπ x}{1}) d x

- \frac{1}{πn} cos (πn x) + C

Solution steps

\int sin (\frac{nπ x}{1}) d x

Simplify

= \int sin (πn x) d x

Apply u-substitution

: \int \frac{sin ( u )}{πn} d u

= \int \frac{sin ( u )}{πn} d u

Take the constant out:

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

= \frac{1}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{πn} (- cos (u))

Substitute back

u = πn x

= \frac{1}{πn} (- cos (πn x))

Simplify

= - \frac{1}{πn} cos (πn x)

Add a constant to the solution

= - \frac{1}{πn} cos (πn x) + C

x \to 1 + lim (x + 1)

x \to 1 lim (\frac{x ^{2} + 1}{x - 2})

\int \frac{cos ( θ )}{1 - sin ( θ )} d θ

y = 6 ln (x^{2}), (e, 12)

f (x) = e^{x} sin (x), (0, 0)

What is the integral of sin((npix)/1) ?
The integral of sin((npix)/1) is -1/(pin)cos(pinx)+C