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

The integral from 0 to pi of sin(n)x is sin(n)(pi^2)/2

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integral from 0 to pi of sin(n)x

\int_{0}^{π} sin (n) x d x

sin (n) \frac{π ^{2}}{2}

Solution steps

\int_{0}^{π} 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} x d x

Apply the Power Rule

: [\frac{x ^{2}}{2}]_{0}^{π}

= sin (n) [\frac{x ^{2}}{2}]_{0}^{π}

Compute the boundaries:

\frac{π ^{2}}{2}

= sin (n) \frac{π ^{2}}{2}

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\int_{- π}^{π} ∣ sin (t) ∣ d t

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

What is the integral from 0 to pi of sin(n)x ?
The integral from 0 to pi of sin(n)x is sin(n)(pi^2)/2