What is the integral from 0 to 2pi of cos^7(x) ?

The integral from 0 to 2pi of cos^7(x) is 0

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integral from 0 to 2pi of cos^7(x)

\int_{0}^{2 π} cos^{7} (x) d x

0

Solution steps

\int_{0}^{2 π} cos^{7} (x) d x

Simplify

cos^{7} (x) : cos^{6} (x) cos (x)

= \int_{0}^{2 π} cos^{6} (x) cos (x) d x

Apply Integration By Parts

: [cos^{6} (x) sin (x) - \int - 6 cos^{5} (x) sin^{2} (x) d x]_{0}^{2 π}

= [cos^{6} (x) sin (x) - \int - 6 cos^{5} (x) sin^{2} (x) d x]_{0}^{2 π}

\int - 6 cos^{5} (x) sin^{2} (x) d x = - 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7})

= [cos^{6} (x) sin (x) - (- 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}))]_{0}^{2 π}

Simplify

[cos^{6} (x) sin (x) - (- 6 (\frac{sin ^{3} ( x )}{3} - \frac{2 sin ^{5} ( x )}{5} + \frac{sin ^{7} ( x )}{7}))]_{0}^{2 π} : [cos^{6} (x) sin (x) + 6 (\frac{1}{3} sin^{3} (x) - \frac{2}{5} sin^{5} (x) + \frac{1}{7} sin^{7} (x))]_{0}^{2 π}

= [cos^{6} (x) sin (x) + 6 (\frac{1}{3} sin^{3} (x) - \frac{2}{5} sin^{5} (x) + \frac{1}{7} sin^{7} (x))]_{0}^{2 π}

Compute the boundaries:

0

= 0

\int_{1}^{1.5} \frac{3}{2 x ^{2}} d x

\int_{1}^{5} 2 \cdot x d x

\int_{3}^{x} sec (t) tan (t) d t

\int_{0}^{1} 2 x (x^{2} + 3)^{4} d x

\int_{0}^{1} \frac{1}{4 x} + 9 d x

What is the integral from 0 to 2pi of cos^7(x) ?
The integral from 0 to 2pi of cos^7(x) is 0