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}^{2} x + 2 - x d x

\int_{2}^{4} \frac{x - 2}{x ^{2} - 5 x + 4} d x

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

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

\int_{0}^{1} (4 - x^{2}) d x

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)