What is the integral from 0 to pi/4 of sec^3(x) ?

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

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integral from 0 to pi/4 of sec^3(x)

\int_{0}^{\frac{π}{4}} sec^{3} (x) d x

\frac{2 + ln ( 1 + 2 )}{2}

Decimal

1.14779 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sec^{3} (x) d x

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

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

Simplify

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

y^{^{'}} + y = cos (x)

f (x) = 4 x^{2} + 2 x - 1, at x = 2

\frac{\partial}{\partial z} (\frac{1}{x y ^{2} z})

\int \frac{x ^{3}}{( 1 + x ^{2} ) ^{3}} d x

x \to \infty lim (\frac{3 x ^{2} - 2 x - 5}{x - 4})

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