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

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

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integral from-pi to 0 of (t+pi/2)cos(nt)

\int_{- π}^{0} (t + \frac{π}{2}) cos (n t) d t

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

Solution steps

\int_{- π}^{0} (t + \frac{π}{2}) cos (n t) d t

Expand

(t + \frac{π}{2}) cos (n t) : t cos (n t) + \frac{π}{2} cos (n t)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

= \frac{1 - ( - 1 ) ^{n}}{n ^{2}} + 0

Simplify

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

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