What is the integral from 0 to pi of 3cos^6(x) ?

The integral from 0 to pi of 3cos^6(x) is (15pi)/(16)

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integral from 0 to pi of 3cos^6(x)

\int_{0}^{π} 3 cos^{6} (x) d x

\frac{15 π}{16}

Decimal Notation

2.94524 \dots

Solution steps

\int_{0}^{π} 3 cos^{6} (x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{π} cos^{6} (x) d x

Apply Integral Reduction

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

\int_{0}^{π} cos^{4} (x) d x = \frac{3 π}{8}

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

Simplify

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

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

Compute the boundaries:

0

= 3 (0 + \frac{5 π}{16})

Simplify

= \frac{15 π}{16}

\int \frac{1}{t ^{2} t ^{2} - 16} d t

\frac{d}{d x} (e^{\frac{x}{6}})

\int \frac{1}{1 + e ^{- t}} d t

\int \frac{x ^{2}}{x ^{2} + 81} d x

What is the integral from 0 to pi of 3cos^6(x) ?
The integral from 0 to pi of 3cos^6(x) is (15pi)/(16)