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

The integral from 0 to pi of x^2cos(nx) is (2pi(-1)^n)/(n^2)

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

\int_{0}^{π} x^{2} cos (n x) d x

\frac{2 π ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{0}^{π} x^{2} cos (n x) d x

Apply u-substitution

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

= \int_{0}^{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_{0}^{nπ} u^{2} cos (u) d u

Apply Integration By Parts

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

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

Compute the boundaries:

2 π (- 1)^{n} n

= \frac{1}{n ^{3}} \cdot 2 π (- 1)^{n} n

Simplify

= \frac{2 π ( - 1 ) ^{n}}{n ^{2}}

\int \frac{1}{x ^{2} x ^{2} - 7} d x

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

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

\frac{d}{d x} (x^{3} + 3 x^{2} - 24 x - 80)

f (x) = x^{3}, at x = 0

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