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

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

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

\int_{0}^{2 π} cos (x) cos (2 x) d x

0

Solution steps

\int_{0}^{2 π} cos (x) cos (2 x) d x

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

0

= 0

\int sin^{3} (x) cos^{8} (x) d x

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

x \to 0 + lim (\frac{1}{3 x})

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

\int \frac{1}{9 - 6 x} d x

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