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

The integral from 0 to pi of sin(x-nx) is ((-1)^{-n}+1)/(1-n)

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integral from 0 to pi of sin(x-nx)

\int_{0}^{π} sin (x - n x) d x

\frac{( - 1 ) ^{- n} + 1}{1 - n}

Solution steps

\int_{0}^{π} sin (x - n x) d x

Apply u-substitution

: \int_{0}^{π - nπ} \frac{sin ( u )}{1 - n} d u

= \int_{0}^{π - nπ} \frac{sin ( u )}{1 - n} d u

Take the constant out:

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

= \frac{1}{1 - n} \cdot \int_{0}^{π - nπ} sin (u) d u

Use the common integral:

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

= \frac{1}{1 - n} [- cos (u)]_{0}^{π - nπ}

Compute the boundaries:

(- 1)^{- n} + 1

= \frac{1}{1 - n} ((- 1)^{- n} + 1)

Simplify

= \frac{( - 1 ) ^{- n} + 1}{1 - n}

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

\int_{1}^{e} \frac{x}{x + 1} d x

\int_{0}^{21 π} sin (x) d x

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

\int_{0}^{2 π} - 150 sin^{2} (t) d t

What is the integral from 0 to pi of sin(x-nx) ?
The integral from 0 to pi of sin(x-nx) is ((-1)^{-n}+1)/(1-n)