What is the integral from 0 to pi/4 of sec^5(x) ?

The integral from 0 to pi/4 of sec^5(x) is (3sqrt(2)ln(1+sqrt(2))+14)/(8sqrt(2))

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integral from 0 to pi/4 of sec^5(x)

\int_{0}^{\frac{π}{4}} sec^{5} (x) d x

\frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

Decimal

1.56795 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sec^{5} (x) d x

Apply Integral Reduction

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

\int_{0}^{\frac{π}{4}} sec^{3} (x) d x = \frac{2 + ln ( 1 + 2 )}{2}

= [\frac{sec ^{4} ( x ) sin ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{2 + ln ( 1 + 2 )}{2}

Simplify

[\frac{sec ^{4} ( x ) sin ( x )}{4}]_{0}^{\frac{π}{4}} + \frac{3}{4} \cdot \frac{2 + ln ( 1 + 2 )}{2} : [\frac{1}{4} sec^{4} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{3 ( 2 + ln ( 1 + 2 ) )}{8}

= [\frac{1}{4} sec^{4} (x) sin (x)]_{0}^{\frac{π}{4}} + \frac{3 ( 2 + ln ( 1 + 2 ) )}{8}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2} + \frac{3 ( 2 + ln ( 1 + 2 ) )}{8}

Simplify

= \frac{3 2 ln ( 1 + 2 ) + 14}{8 2}

f (x) = 8 x^{3} - x^{π}

\int \frac{x}{- 5 x ^{2} + 2 x + 1} d x

\frac{d}{d x} (8 (x + 4)^{\frac{3}{2}})

x \to - 2 lim (\frac{x + 1}{1 - x ^{2}})

\frac{d}{d x} (r (x) cos (x))

What is the integral from 0 to pi/4 of sec^5(x) ?
The integral from 0 to pi/4 of sec^5(x) is (3sqrt(2)ln(1+sqrt(2))+14)/(8sqrt(2))