What is the integral from 0 to pi of 11cos^6(x) ?

The integral from 0 to pi of 11cos^6(x) is (55pi)/(16)

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integral from 0 to pi of 11cos^6(x)

\int_{0}^{π} 11 cos^{6} (x) d x

\frac{55 π}{16}

Decimal

10.79922 \dots

Solution steps

\int_{0}^{π} 11 cos^{6} (x) d x

Take the constant out:

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

= 11 \cdot \int_{0}^{π} cos^{6} (x) d x

Apply Integral Reduction

= 11 ([\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}

= 11 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8})

Simplify

11 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8}) : 11 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

= 11 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

Compute the boundaries:

0

= 11 (0 + \frac{5 π}{16})

Simplify

= \frac{55 π}{16}

n = 2 \sum \infty ln (n - 1)

5 (3 t + 4)^{2}

\frac{d}{d x} (x^{2} + 5 x + 6)

y^{^{'}} - (\frac{2}{x} y) = y^{5} x^{- 3}

x^{3} ln (4 x)

What is the integral from 0 to pi of 11cos^6(x) ?
The integral from 0 to pi of 11cos^6(x) is (55pi)/(16)