What is the integral from-pi to 0 of (pi+t)cos(nt) ?

The integral from-pi to 0 of (pi+t)cos(nt) is (1-(-1)^n)/(n^2)

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integral from-pi to 0 of (pi+t)cos(nt)

\int_{- π}^{0} (π + t) cos (n t) d t

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

Solution steps

\int_{- π}^{0} (π + t) cos (n t) d t

Apply Integration By Parts

: [\frac{1}{n} sin (n t) (π + t) - \int \frac{1}{n} sin (n t) d t]_{- π}^{0}

= [\frac{1}{n} sin (n t) (π + t) - \int \frac{1}{n} sin (n t) d t]_{- π}^{0}

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

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

Simplify

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

Compute the boundaries:

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

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

Simplify

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

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What is the integral from-pi to 0 of (pi+t)cos(nt) ?
The integral from-pi to 0 of (pi+t)cos(nt) is (1-(-1)^n)/(n^2)