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

The integral from-pi to pi of x^2cos(nx) is (4pi(-1)^n)/(n^2)

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

\int_{- π}^{π} x^{2} cos (n x) d x

\frac{4 π ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{- π}^{π} x^{2} cos (n x) d x

Apply u-substitution

: \int_{- πn}^{nπ} \frac{u ^{2} cos ( u )}{n ^{3}} d u

= \int_{- πn}^{nπ} \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_{- πn}^{nπ} u^{2} cos (u) d u

Apply Integration By Parts

: [u^{2} sin (u) - \int 2 u sin (u) d u]_{- πn}^{nπ}

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

\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))]_{- πn}^{nπ}

Compute the boundaries:

4 π (- 1)^{n} n

= \frac{1}{n ^{3}} \cdot 4 π (- 1)^{n} n

Simplify

= \frac{4 π ( - 1 ) ^{n}}{n ^{2}}

\int_{0}^{4} π (x)^{2} d x

y^{^{'}} + 3 y = 15 - e^{t}

x \to 5 - lim (e^{\frac{3}{( 5 - x )}})

\frac{d}{d x} (csc^{7} (x))

y = (x^{2} - 35)^{7}, at x = 6

What is the integral from-pi to pi of x^2cos(nx) ?
The integral from-pi to pi of x^2cos(nx) is (4pi(-1)^n)/(n^2)