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

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

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

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

Solution steps

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

Apply Integration By Parts

: [\frac{1}{n} (- cos (n x) sin^{2} (x) - n \cdot \int - \frac{1}{n} cos (n x) sin (2 x) d x)]_{0}^{π}

= [\frac{1}{n} (- cos (n x) sin^{2} (x) - n \cdot \int - \frac{1}{n} cos (n x) sin (2 x) d x)]_{0}^{π}

\int - \frac{1}{n} cos (n x) sin (2 x) d x = - \frac{1}{2 n} (- \frac{1}{2 + n} cos (2 x + n x) - \frac{1}{2 - n} cos (2 x - n x))

= [\frac{1}{n} (- cos (n x) sin^{2} (x) - n (- \frac{1}{2 n} (- \frac{1}{2 + n} cos (2 x + n x) - \frac{1}{2 - n} cos (2 x - n x))))]_{0}^{π}

Simplify

[\frac{1}{n} (- cos (n x) sin^{2} (x) - n (- \frac{1}{2 n} (- \frac{1}{2 + n} cos (2 x + n x) - \frac{1}{2 - n} cos (2 x - n x))))]_{0}^{π} : [\frac{1}{n} (- cos (n x) sin^{2} (x) - \frac{1}{2 ( 2 + n )} cos (2 x + n x) - \frac{1}{2 ( 2 - n )} cos (2 x - n x))]_{0}^{π}

= [\frac{1}{n} (- cos (n x) sin^{2} (x) - \frac{1}{2 ( 2 + n )} cos (2 x + n x) - \frac{1}{2 ( 2 - n )} cos (2 x - n x))]_{0}^{π}

Compute the boundaries:

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

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

Simplify

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

\int_{0}^{t} sin (t - x) d x

\int_{0}^{1} x^{2} ln (1 + x^{2}) d x

\int_{1}^{3} x^{2} - 2 x + 5 d x

\int_{0}^{π} \frac{1}{2} sin (x^{2}) d x

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

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