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

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

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

\int_{0}^{2 π} (cos (x))^{4} d x

\frac{3 π}{4}

Solution steps

\int_{0}^{2 π} (cos (x))^{4} d x

Simplify

= \int_{0}^{2 π} cos^{4} (x) d x

Simplify

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

= \int_{0}^{2 π} cos^{3} (x) cos (x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{3 π}{4}

= \frac{3 π}{4}

\int_{- 2}^{0} \frac{1}{( 2 - 1 ) ^{3}} d x

\int_{- 7}^{3} ∣ x ∣ d x

\int_{0}^{1} arcsin (x) d x

\int_{0}^{1} 4 π x d x

\int_{3 y}^{3} e^{x^{2}} d x

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