What is the integral from 0 to pi/2 of cos^8(x) ?

The integral from 0 to pi/2 of cos^8(x) is (35pi)/(256)

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integral from 0 to pi/2 of cos^8(x)

\int_{0}^{\frac{π}{2}} cos^{8} (x) d x

\frac{35 π}{256}

Solution steps

\int_{0}^{\frac{π}{2}} cos^{8} (x) d x

Apply Integral Reduction

= [\frac{sin ( x ) cos ^{7} ( x )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \int_{0}^{\frac{π}{2}} cos^{6} (x) d x

\int_{0}^{\frac{π}{2}} cos^{6} (x) d x = \frac{5 π}{32}

= [\frac{sin ( x ) cos ^{7} ( x )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32}

Simplify

[\frac{sin ( x ) cos ^{7} ( x )}{8}]_{0}^{\frac{π}{2}} + \frac{7}{8} \cdot \frac{5 π}{32} : [\frac{1}{8} cos^{7} (x) sin (x)]_{0}^{\frac{π}{2}} + \frac{35 π}{256}

= [\frac{1}{8} cos^{7} (x) sin (x)]_{0}^{\frac{π}{2}} + \frac{35 π}{256}

Compute the boundaries:

0

= 0 + \frac{35 π}{256}

Simplify

= \frac{35 π}{256}

f (t) = t^{3} - 9 t^{2} + 24 t

\int \frac{x}{( x ^{3} - 1 )} d x

y^{^{''}} + 16 y = 3 sin (8 t), y (0) = 0, y^{^{'}} (0) = 0

\frac{\partial}{\partial y} (\frac{3 x - 3 y}{3 x + 3 y})

x \to 0 lim (x - \frac{1}{x - 1})

What is the integral from 0 to pi/2 of cos^8(x) ?
The integral from 0 to pi/2 of cos^8(x) is (35pi)/(256)