What is the integral from-1 to 0 of tcos(npit) ?

The integral from-1 to 0 of tcos(npit) is (1-(-1)^n)/(pi^2n^2)

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integral from-1 to 0 of tcos(npit)

\int_{- 1}^{0} t cos (nπ t) d t

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

Solution steps

\int_{- 1}^{0} t cos (nπ t) d t

Apply u-substitution

: \int_{- πn}^{0} \frac{u cos ( u )}{π ^{2} n ^{2}} d u

= \int_{- πn}^{0} \frac{u cos ( u )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int_{- πn}^{0} u cos (u) d u

Apply Integration By Parts

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

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

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

= \frac{1}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{- πn}^{0}

Simplify

= \frac{1}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{- πn}^{0}

Compute the boundaries:

1 - (- 1)^{n}

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

Simplify

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

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What is the integral from-1 to 0 of tcos(npit) ?
The integral from-1 to 0 of tcos(npit) is (1-(-1)^n)/(pi^2n^2)