What is the integral from 0 to 2pi of cos^3(t) ?

The integral from 0 to 2pi of cos^3(t) is 0

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integral from 0 to 2pi of cos^3(t)

\int_{0}^{2 π} cos^{3} (t) d t

0

Solution steps

\int_{0}^{2 π} cos^{3} (t) d t

Simplify

cos^{3} (t) : cos^{2} (t) cos (t)

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

Rewrite using trig identities

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

Apply Integration By Parts

: [- \int - sin (2 t) sin (t) d t + cos^{2} (t) sin (t)]_{0}^{2 π}

= [- \int - sin (2 t) sin (t) d t + cos^{2} (t) sin (t)]_{0}^{2 π}

\int - sin (2 t) sin (t) d t = - \frac{2}{3} sin^{3} (t)

= [- (- \frac{2}{3} sin^{3} (t)) + cos^{2} (t) sin (t)]_{0}^{2 π}

Simplify

[- (- \frac{2}{3} sin^{3} (t)) + cos^{2} (t) sin (t)]_{0}^{2 π} : [\frac{1}{3} (2 sin^{3} (t) + 3 cos^{2} (t) sin (t))]_{0}^{2 π}

= [\frac{1}{3} (2 sin^{3} (t) + 3 cos^{2} (t) sin (t))]_{0}^{2 π}

Compute the boundaries:

0

= 0

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What is the integral from 0 to 2pi of cos^3(t) ?
The integral from 0 to 2pi of cos^3(t) is 0