What is the integral of sin(4pix)sin(npix) ?

The integral of sin(4pix)sin(npix) is 1/2 (-1/(4pi+pin)sin(4pix+npix)+1/(4pi-pin)sin(4pix-npix))+C

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integral of sin(4pix)sin(npix)

\int sin (4 π x) sin (nπ x) d x

\frac{1}{2} (- \frac{1}{4 π + πn} sin (4 π x + nπ x) + \frac{1}{4 π - πn} sin (4 π x - nπ x)) + C

Solution steps

\int sin (4 π x) sin (nπ x) d x

Rewrite using trig identities

= \int \frac{- cos ( 4 π x + nπ x ) + cos ( 4 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 (4 π x + nπ x) + cos (4 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 (4 π x + nπ x) d x + \int cos (4 π x - nπ x) d x)

\int cos (4 π x + nπ x) d x = \frac{1}{4 π + πn} sin (4 π x + nπ x)

\int cos (4 π x - nπ x) d x = \frac{1}{4 π - πn} sin (4 π x - nπ x)

= \frac{1}{2} (- \frac{1}{4 π + πn} sin (4 π x + nπ x) + \frac{1}{4 π - πn} sin (4 π x - nπ x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4 π + πn} sin (4 π x + nπ x) + \frac{1}{4 π - πn} sin (4 π x - nπ x)) + C

\frac{d}{d x} (lo g_{3} (8 x - 1))

\int_{0}^{1} 5^{- θ} d θ

\frac{d y}{d x} = - 2 x y^{2}

\frac{d y}{d x} = \frac{x ^{8} - 4 y}{x}

\int (1 - a x)^{\frac{1}{2}} d x

What is the integral of sin(4pix)sin(npix) ?
The integral of sin(4pix)sin(npix) is 1/2 (-1/(4pi+pin)sin(4pix+npix)+1/(4pi-pin)sin(4pix-npix))+C