What is the integral from 0 to pi of cos(x)*sin(nx) ?

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

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 cos(x)*sin(nx)

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

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

Solution steps

\int_{0}^{π} cos (x) sin (n x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{π} sin (n x + x) d x + \int_{0}^{π} sin (n x - x) d x)

\int_{0}^{π} sin (n x + x) d x = \frac{( - 1 ) ^{n} + 1}{n + 1}

\int_{0}^{π} sin (n x - x) d x = \frac{( - 1 ) ^{n} + 1}{n - 1}

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

Simplify

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

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

\int_{0}^{1} 2 t^{2} d t

\int_{1}^{9} \frac{e ^{\frac{9}{x}}}{x ^{2}} d x

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

\int_{2}^{5} - \frac{3}{x ^{2}} d x

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

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