What is the integral from 0 to pi/4 of cos^6(x) ?

The integral from 0 to pi/4 of cos^6(x) is 1/48+(5(3pi+8))/(192)

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integral from 0 to pi/4 of cos^6(x)

\int_{0}^{\frac{π}{4}} cos^{6} (x) d x

\frac{1}{48} + \frac{5 ( 3 π + 8 )}{192}

Decimal

0.47460 \dots

Solution steps

\int_{0}^{\frac{π}{4}} cos^{6} (x) d x

Apply Integral Reduction

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

\int_{0}^{\frac{π}{4}} cos^{4} (x) d x = \frac{1}{16} + \frac{3 ( π + 2 )}{32}

= [\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{\frac{π}{4}} + \frac{5}{6} (\frac{1}{16} + \frac{3 ( π + 2 )}{32})

Simplify

[\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{\frac{π}{4}} + \frac{5}{6} (\frac{1}{16} + \frac{3 ( π + 2 )}{32}) : [\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{5 ( 3 π + 8 )}{192}

= [\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{5 ( 3 π + 8 )}{192}

Compute the boundaries:

\frac{1}{48}

= \frac{1}{48} + \frac{5 ( 3 π + 8 )}{192}

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What is the integral from 0 to pi/4 of cos^6(x) ?
The integral from 0 to pi/4 of cos^6(x) is 1/48+(5(3pi+8))/(192)