What is the integral from 0 to pi/2 of cos^5(6x) ?

The integral from 0 to pi/2 of cos^5(6x) is 0

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integral from 0 to pi/2 of cos^5(6x)

\int_{0}^{\frac{π}{2}} cos^{5} (6 x) d x

0

Solution steps

\int_{0}^{\frac{π}{2}} cos^{5} (6 x) d x

Apply u-substitution

: \int_{0}^{3 π} cos^{5} (u) \frac{1}{6} d u

= \int_{0}^{3 π} cos^{5} (u) \frac{1}{6} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{0}^{3 π} cos^{5} (u) d u

Apply Integral Reduction

= \frac{1}{6} ([\frac{sin ( u ) cos ^{4} ( u )}{5}]_{0}^{3 π} + \frac{4}{5} \cdot \int_{0}^{3 π} cos^{3} (u) d u)

\int_{0}^{3 π} cos^{3} (u) d u = 0

= \frac{1}{6} ([\frac{sin ( u ) cos ^{4} ( u )}{5}]_{0}^{3 π} + \frac{4}{5} \cdot 0)

Simplify

\frac{1}{6} ([\frac{sin ( u ) cos ^{4} ( u )}{5}]_{0}^{3 π} + \frac{4}{5} \cdot 0) : \frac{1}{6} [\frac{1}{5} cos^{4} (u) sin (u)]_{0}^{3 π}

= \frac{1}{6} [\frac{1}{5} cos^{4} (u) sin (u)]_{0}^{3 π}

Compute the boundaries:

0

= \frac{1}{6} \cdot 0

Simplify

= 0

\int \frac{sec ^{2} ( x )}{1 + tan ^{2} ( x )} d x

\frac{d}{d x} (arctan (x) + 2 x - 2)

y = x^{\frac{15}{16}}

\int 2 x - \frac{1}{2 x} d x

\int \frac{1}{y + 1} d y

What is the integral from 0 to pi/2 of cos^5(6x) ?
The integral from 0 to pi/2 of cos^5(6x) is 0