What is the integral from 0 to pi of 13cos^6(x) ?

The integral from 0 to pi of 13cos^6(x) is (65pi)/(16)

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integral from 0 to pi of 13cos^6(x)

\int_{0}^{π} 13 cos^{6} (x) d x

\frac{65 π}{16}

Decimal Notation

12.76272 \dots

Solution steps

\int_{0}^{π} 13 cos^{6} (x) d x

Take the constant out:

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

= 13 \cdot \int_{0}^{π} cos^{6} (x) d x

Apply Integral Reduction

= 13 ([\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}

= 13 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8})

Simplify

13 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8}) : 13 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

= 13 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

Compute the boundaries:

0

= 13 (0 + \frac{5 π}{16})

Simplify

= \frac{65 π}{16}

\frac{d y}{d x} = \frac{x}{3 y}

x \to 0 + lim (\frac{x ^{- 2}}{2})

\int sin (x) cos^{2} (d) x d x

\frac{d}{d x} ((- 3)^{x})

(4, - 3), (- 2, 7)

What is the integral from 0 to pi of 13cos^6(x) ?
The integral from 0 to pi of 13cos^6(x) is (65pi)/(16)