What is the integral from 0 to 2pi of 1/3 cos^4(t) ?

The integral from 0 to 2pi of 1/3 cos^4(t) is pi/4

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integral from 0 to 2pi of 1/3 cos^4(t)

\int_{0}^{2 π} \frac{1}{3} cos^{4} (t) d t

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{0}^{2 π} \frac{1}{3} cos^{4} (t) d t

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{2 π} cos^{4} (t) d t

Simplify

cos^{4} (t) : cos^{3} (t) cos (t)

= \frac{1}{3} \cdot \int_{0}^{2 π} cos^{3} (t) cos (t) d t

Apply Integration By Parts

: [cos^{3} (t) sin (t) - \int - 3 cos^{2} (t) sin^{2} (t) d t]_{0}^{2 π}

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

\int - 3 cos^{2} (t) sin^{2} (t) d t = - \frac{3}{8} (t - \frac{1}{4} sin (4 t))

= \frac{1}{3} [cos^{3} (t) sin (t) - (- \frac{3}{8} (t - \frac{1}{4} sin (4 t)))]_{0}^{2 π}

Simplify

\frac{1}{3} [cos^{3} (t) sin (t) - (- \frac{3}{8} (t - \frac{1}{4} sin (4 t)))]_{0}^{2 π} : \frac{1}{3} [\frac{1}{32} (32 cos^{3} (t) sin (t) + 3 (4 t - sin (4 t)))]_{0}^{2 π}

= \frac{1}{3} [\frac{1}{32} (32 cos^{3} (t) sin (t) + 3 (4 t - sin (4 t)))]_{0}^{2 π}

Compute the boundaries:

\frac{3 π}{4}

= \frac{1}{3} \cdot \frac{3 π}{4}

Simplify

= \frac{π}{4}

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What is the integral from 0 to 2pi of 1/3 cos^4(t) ?
The integral from 0 to 2pi of 1/3 cos^4(t) is pi/4