What is the integral from-1 to 1 of t*sin(npit) ?

The integral from-1 to 1 of t*sin(npit) is -(2(-1)^n)/(pin)

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integral from-1 to 1 of t*sin(npit)

\int_{- 1}^{1} t \cdot sin (nπ t) d t

- \frac{2 ( - 1 ) ^{n}}{πn}

Solution steps

\int_{- 1}^{1} t sin (nπ t) d t

Apply Integration By Parts

= [- \frac{1}{πn} t cos (πn t) - \int - \frac{1}{πn} cos (πn t) d t]_{- 1}^{1}

\int - \frac{1}{πn} cos (πn t) d t = - \frac{1}{π ^{2} n ^{2}} sin (πn t)

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

Simplify

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

Compute the boundaries:

- (- 1)^{n} \frac{2}{πn}

= - (- 1)^{n} \frac{2}{πn}

Simplify

= - \frac{2 ( - 1 ) ^{n}}{πn}

What is the integral from-1 to 1 of t*sin(npit) ?
The integral from-1 to 1 of t*sin(npit) is -(2(-1)^n)/(pin)