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

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

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

\int_{0}^{π} x^{2} cos (k x) d x

\frac{2 π ( - 1 ) ^{k}}{k ^{2}}

Solution steps

\int_{0}^{π} x^{2} cos (k x) d x

Apply u-substitution

: \int_{0}^{kπ} \frac{u ^{2} cos ( u )}{k ^{3}} d u

= \int_{0}^{kπ} \frac{u ^{2} cos ( u )}{k ^{3}} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

2 π (- 1)^{k} k

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

Simplify

= \frac{2 π ( - 1 ) ^{k}}{k ^{2}}

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