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

The integral from 0 to 2pi of x^2cos(nx) is (4pi^2n^2sin(2pin)-2(sin(2pin)-2pincos(2pin)))/(n^3)

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integral from 0 to 2pi of x^2cos(nx)

\int_{0}^{2 π} x^{2} cos (n x) d x

\frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{n ^{3}}

Solution steps

\int_{0}^{2 π} x^{2} cos (n x) d x

Apply u-substitution

: \int_{0}^{n \cdot 2 π} \frac{u ^{2} cos ( u )}{n ^{3}} d u

= \int_{0}^{n \cdot 2 π} \frac{u ^{2} cos ( u )}{n ^{3}} d u

Take the constant out:

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

= \frac{1}{n ^{3}} \cdot \int_{0}^{n \cdot 2 π} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{n \cdot 2 π}

= \frac{1}{n ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{n \cdot 2 π}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

= \frac{1}{n ^{3}} [u^{2} sin (u) - 2 (- u cos (u) + sin (u))]_{0}^{n \cdot 2 π}

Compute the boundaries:

4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn))

= \frac{1}{n ^{3}} (4 π^{2} n^{2} sin (2 πn) - 2 (sin (2 πn) - 2 πn cos (2 πn)))

Simplify

= \frac{4 π ^{2} n ^{2} sin ( 2 πn ) - 2 ( sin ( 2 πn ) - 2 πn cos ( 2 πn ) )}{n ^{3}}

\int \frac{ln ( 22 x )}{x} d x

\int x^{3} 1 + 9 x^{4} d x

\frac{- 3 s - 4}{2 ( s ^{2} + 2 s + 2 )}

\frac{e ^{- 8 s}}{s}

\frac{d}{d x} (e^{2 e^{x}})

What is the integral from 0 to 2pi of x^2cos(nx) ?
The integral from 0 to 2pi of x^2cos(nx) is (4pi^2n^2sin(2pin)-2(sin(2pin)-2pincos(2pin)))/(n^3)