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

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

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

\int_{0}^{2 π} cos^{3} (x) d x

0

Solution steps

\int_{0}^{2 π} cos^{3} (x) d x

Simplify

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

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

Rewrite using trig identities

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

0

= 0

3 x y y^{^{'}} + x^{2} + y^{2} = 0

y^{^{''}} - 2 y^{^{'}} = x

\int \frac{1}{9 + 4 x ^{2}} d x

\frac{d}{d x} (\frac{9}{2 x + 1})

\int_{1 - 3 x}^{1} \frac{u ^{3}}{1 + u ^{2}} d u

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