What is the integral from 0 to pi of 2sin(nx) ?

The integral from 0 to pi of 2sin(nx) is (2(-(-1)^n+1))/n

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integral from 0 to pi of 2sin(nx)

\int_{0}^{π} 2 sin (n x) d x

\frac{2 ( - ( - 1 ) ^{n} + 1 )}{n}

Solution steps

\int_{0}^{π} 2 sin (n x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{π} sin (n x) d x

Apply u-substitution

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

= 2 \cdot \int_{0}^{nπ} \frac{sin ( u )}{n} d u

Take the constant out:

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

= 2 \cdot \frac{1}{n} \cdot \int_{0}^{nπ} sin (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{n} [- cos (u)]_{0}^{nπ}

Simplify

2 \cdot \frac{1}{n} [- cos (u)]_{0}^{nπ} : \frac{2}{n} [- cos (u)]_{0}^{πn}

= \frac{2}{n} [- cos (u)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{2}{n} (- (- 1)^{n} + 1)

Simplify

= \frac{2 ( - ( - 1 ) ^{n} + 1 )}{n}

f (x) = 14 x

\frac{d}{d x} (2 arctan (e^{x}))

\frac{d}{d x} (8 (x + 2)^{\frac{3}{2}})

\frac{d}{d x} (- 2 x^{4} + 3 x^{3} - x^{2})

\frac{d y}{d x}, 40 x^{\frac{1}{3}} y^{\frac{2}{3}} = 480

What is the integral from 0 to pi of 2sin(nx) ?
The integral from 0 to pi of 2sin(nx) is (2(-(-1)^n+1))/n