What is the integral from-pi to pi of t^2cos(nt) ?

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

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integral from-pi to pi of t^2cos(nt)

\int_{- π}^{π} t^{2} cos (n t) d t

\frac{4 π ( - 1 ) ^{n}}{n ^{2}}

Solution steps

\int_{- π}^{π} t^{2} cos (n t) d t

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_{1}^{2} \frac{x}{y} + \frac{y}{x} d y

\int_{1}^{2 x} \frac{1}{1 + t ^{3}} d t

\int_{0}^{2} 2 x 12 - x^{2} d x

\int_{0}^{a x} e^{- t^{2}} d t

\int_{0}^{t} u e^{t - u} d u

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