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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{- π}^{0} \frac{π}{2} sin (n x) d x

\frac{π ( - 1 + ( - 1 ) ^{n} )}{2 n}

Solution steps

\int_{- π}^{0} \frac{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

= \frac{π}{2} \cdot \int_{- π}^{0} sin (n x) d x

Apply u-substitution

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

= \frac{π}{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

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

Use the common integral:

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

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

Simplify

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

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

Compute the boundaries:

- 1 + (- 1)^{n}

= \frac{π}{2 n} (- 1 + (- 1)^{n})

Simplify

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

\int_{0}^{1} 3 x^{2} + 2 x + 1 d x

\int_{0}^{\frac{π}{2}} sin (2 y) d y

\int_{\frac{1}{3}}^{3} 6 - \frac{6}{x ^{2}} d x

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

\int_{0}^{1} 6 x d x

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