What is the integral from 0 to pi of sin(t)sin(nt) ?

The integral from 0 to pi of sin(t)sin(nt) is 0

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integral from 0 to pi of sin(t)sin(nt)

\int_{0}^{π} sin (t) sin (n t) d t

0

Solution steps

\int_{0}^{π} sin (t) sin (n t) d t

Rewrite using trig identities

= \int_{0}^{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_{0}^{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_{0}^{π} cos (t + n t) d t + \int_{0}^{π} cos (t - n t) d t)

\int_{0}^{π} cos (t + n t) d t = 0

\int_{0}^{π} cos (t - n t) d t = 0

= \frac{1}{2} (- 0 + 0)

Simplify

= 0

\int_{0}^{3} (\frac{1}{x})^{2} d x

\int_{- 3}^{5} ∣ x + 1 ∣ d x

\int_{0}^{3} (2) d x

\int_{0}^{1} \frac{x}{1 + 2 x ^{2}} d x

\int_{2}^{4} f (t) d t

What is the integral from 0 to pi of sin(t)sin(nt) ?
The integral from 0 to pi of sin(t)sin(nt) is 0