What is the integral from 0 to pi of 9cos^6(x) ?

The integral from 0 to pi of 9cos^6(x) is (45pi)/(16)

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integral from 0 to pi of 9cos^6(x)

\int_{0}^{π} 9 cos^{6} (x) d x

\frac{45 π}{16}

Decimal Notation

8.83572 \dots

Solution steps

\int_{0}^{π} 9 cos^{6} (x) d x

Take the constant out:

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

= 9 \cdot \int_{0}^{π} cos^{6} (x) d x

Apply Integral Reduction

= 9 ([\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}

= 9 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8})

Simplify

9 ([\frac{sin ( x ) cos ^{5} ( x )}{6}]_{0}^{π} + \frac{5}{6} \cdot \frac{3 π}{8}) : 9 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

= 9 ([\frac{1}{6} cos^{5} (x) sin (x)]_{0}^{π} + \frac{5 π}{16})

Compute the boundaries:

0

= 9 (0 + \frac{5 π}{16})

Simplify

= \frac{45 π}{16}

\int sin (x^{2} - 1) d x

\frac{d}{d x} (\frac{e ^{x}}{1 - e ^{x}})

\int \frac{5 x ^{2} - x + 17}{( x + 2 ) ( x ^{2} + 9 )} d x

\frac{d y}{d x} = e^{8 x + y}

f (x) = csc (x + 5)

What is the integral from 0 to pi of 9cos^6(x) ?
The integral from 0 to pi of 9cos^6(x) is (45pi)/(16)