What is the integral of sin((npi)/2 x) ?

The integral of sin((npi)/2 x) is -2/(pin)cos((pin)/2 x)+C

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integral of sin((npi)/2 x)

\int sin (\frac{nπ}{2} x) d x

- \frac{2}{πn} cos (\frac{πn}{2} x) + C

Solution steps

\int sin (\frac{nπ}{2} x) d x

Apply u-substitution

: \int \frac{2 sin ( u )}{πn} d u

= \int \frac{2 sin ( u )}{πn} d u

Take the constant out:

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

= \frac{2}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{2}{πn} (- cos (u))

Substitute back

u = \frac{nπ}{2} x

= \frac{2}{πn} (- cos (\frac{nπ}{2} x))

Simplify

= - \frac{2}{πn} cos (\frac{πn}{2} x)

Add a constant to the solution

= - \frac{2}{πn} cos (\frac{πn}{2} x) + C

ln (2 x^{2} + 1)

\frac{d}{d x} (\frac{3}{x - 8})

\int \frac{1}{sin ^{2} ( 5 x )} d x

\frac{d}{d z} (x y z)

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

What is the integral of sin((npi)/2 x) ?
The integral of sin((npi)/2 x) is -2/(pin)cos((pin)/2 x)+C