What is the integral from pi to 2pi of sin(t)sin(nt) ?

The integral from pi to 2pi of sin(t)sin(nt) is 1/2 (-(sin(2pi+2pin))/(1+n)+(sin(2pi-2pin))/(1-n))

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integral from pi to 2pi of sin(t)sin(nt)

\int_{π}^{2 π} sin (t) sin (n t) d t

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

Solution steps

\int_{π}^{2 π} sin (t) sin (n t) d t

Rewrite using trig identities

= \int_{π}^{2 π} \frac{- cos ( t + n t ) + cos ( t - n t )}{2} d t

Take the constant out:

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

= \frac{1}{2} \cdot \int_{π}^{2 π} - cos (t + n t) + cos (t - n t) d t

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_{π}^{2 π} cos (t + n t) d t + \int_{π}^{2 π} cos (t - n t) d t)

\int_{π}^{2 π} cos (t + n t) d t = \frac{sin ( 2 π + 2 πn )}{1 + n}

\int_{π}^{2 π} cos (t - n t) d t = \frac{sin ( 2 π - 2 πn )}{1 - n}

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

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What is the integral from pi to 2pi of sin(t)sin(nt) ?
The integral from pi to 2pi of sin(t)sin(nt) is 1/2 (-(sin(2pi+2pin))/(1+n)+(sin(2pi-2pin))/(1-n))