What is the integral of sin(2pix)sin(npix) ?

The integral of sin(2pix)sin(npix) is 1/2 (-1/(2pi+pin)sin(2pix+npix)+1/(pi(-n+2))sin(2pix-pinx))+C

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integral of sin(2pix)sin(npix)

\int sin (2 π x) sin (nπ x) d x

\frac{1}{2} (- \frac{1}{2 π + πn} sin (2 π x + nπ x) + \frac{1}{π ( - n + 2 )} sin (2 π x - πn x)) + C

Solution steps

\int sin (2 π x) sin (nπ x) d x

Rewrite using trig identities

= \int \frac{- cos ( 2 π x + nπ x ) + cos ( 2 π x - nπ 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 - cos (2 π x + nπ x) + cos (2 π x - nπ x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int cos (2 π x + nπ x) d x + \int cos (2 π x - nπ x) d x)

\int cos (2 π x + nπ x) d x = \frac{1}{2 π + πn} sin (2 π x + nπ x)

\int cos (2 π x - nπ x) d x = \frac{1}{π ( - n + 2 )} sin (2 π x - πn x)

= \frac{1}{2} (- \frac{1}{2 π + πn} sin (2 π x + nπ x) + \frac{1}{π ( - n + 2 )} sin (2 π x - πn x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{2 π + πn} sin (2 π x + nπ x) + \frac{1}{π ( - n + 2 )} sin (2 π x - πn x)) + C

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What is the integral of sin(2pix)sin(npix) ?
The integral of sin(2pix)sin(npix) is 1/2 (-1/(2pi+pin)sin(2pix+npix)+1/(pi(-n+2))sin(2pix-pinx))+C