What is the integral from pi to 2pi of sin(x)sin(nx) ?

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

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integral from pi to 2pi of sin(x)sin(nx)

\int_{π}^{2 π} sin (x) sin (n x) d x

\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 (x) sin (n x) d x

Rewrite using trig identities

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

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

\int_{π}^{2 π} cos (x - n x) d x = \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})

\int_{0}^{5} - 3 x d x

\int_{0}^{8} ∣ x - 4 ∣ d x

\int_{- 2}^{5} 1 d x

\int_{1}^{2} (2 x + 3) d x

\int_{0}^{3} x^{2} - 1 d x

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