What is the integral from-1 to 1 of t^2cos(npit) ?

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

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integral from-1 to 1 of t^2cos(npit)

\int_{- 1}^{1} t^{2} cos (nπ t) d t

\frac{4 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

Solution steps

\int_{- 1}^{1} t^{2} cos (nπ t) d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Compute the boundaries:

4 π (- 1)^{n} n

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

Simplify

= \frac{4 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

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