What is the integral from 0 to pi of sin(x)sin(n)x ?

The integral from 0 to pi of sin(x)sin(n)x is sin(n)pi

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integral from 0 to pi of sin(x)sin(n)x

\int_{0}^{π} sin (x) sin (n) x d x

sin (n) π

Solution steps

\int_{0}^{π} sin (x) sin (n) x d x

Take the constant out:

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

= sin (n) \cdot \int_{0}^{π} sin (x) x d x

Apply Integration By Parts

: [- x cos (x) - \int - cos (x) d x]_{0}^{π}

= sin (n) [- x cos (x) - \int - cos (x) d x]_{0}^{π}

\int - cos (x) d x = - sin (x)

= sin (n) [- x cos (x) - (- sin (x))]_{0}^{π}

Simplify

= sin (n) [- x cos (x) + sin (x)]_{0}^{π}

Compute the boundaries:

π

= sin (n) π

\int_{0}^{1} 14 x e^{x} d x

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

\int_{- \infty}^{- 3} θ e^{θ} d θ

\int_{0}^{5} \frac{13}{x + 4} d x

\int_{0}^{\frac{3 π}{4}} \frac{1}{1 + 9 x ^{2}} d x

What is the integral from 0 to pi of sin(x)sin(n)x ?
The integral from 0 to pi of sin(x)sin(n)x is sin(n)pi