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

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

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

\int_{0}^{2 π} cos^{4} (t) d t

\frac{3 π}{4}

Decimal Notation

2.35619 \dots

Solution steps

\int_{0}^{2 π} cos^{4} (t) d t

Simplify

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

= \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}

= [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))

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

Simplify

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

= [\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{3 π}{4}

\int_{0}^{\frac{π}{4}} cos (2 t) d t

\int_{- \frac{π}{2}}^{\frac{π}{2}} (cos (x) - 5) d x

\int_{0}^{3} (2 - x)^{2} d x

\int_{- 2}^{1} (\frac{1}{x ^{2}} - 5) d x

\int_{3}^{8} \frac{1}{x ^{2}} d x

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