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

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

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integral from-pi to 0 of 2sin(nx)

\int_{- π}^{0} 2 sin (n x) d x

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

Solution steps

\int_{- π}^{0} 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_{- π}^{0} sin (n x) d x

Apply u-substitution

= 2 \cdot \int_{- πn}^{0} \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_{- πn}^{0} sin (u) d u

Use the common integral:

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

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

Simplify

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

= \frac{2}{n} [- cos (u)]_{- πn}^{0}

Compute the boundaries:

- 1 + (- 1)^{n}

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

Simplify

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

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