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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of asin(nx)

\int_{0}^{π} a sin (n x) d x

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

Solution steps

\int_{0}^{π} a sin (n x) d x

Take the constant out:

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

= a \cdot \int_{0}^{π} sin (n x) d x

Apply u-substitution

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

= a \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

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

Use the common integral:

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

= a \frac{1}{n} [- cos (u)]_{0}^{nπ}

Compute the boundaries:

- (- 1)^{n} + 1

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

Simplify

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

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

\int_{1}^{8} (x^{- \frac{10}{3}}) d x

\int_{0}^{1} (t) d t

\int_{0}^{5} \frac{2 x}{x ^{2} + 1} d x

\int_{1}^{4} 3 t d t

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