What is the integral from 0 to pi/4 of 2sec^3(x) ?

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

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integral from 0 to pi/4 of 2sec^3(x)

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

2 + ln (1 + 2)

Decimal Notation

2.29558 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

= 2 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

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

Simplify

= 2 + ln (1 + 2)

(\frac{u ^{3} - 1}{u ^{3} + 1})^{8}

\frac{\partial}{\partial x} (4 - 2 xx)

\frac{1}{s ^{2} + 2 s + 1}

\int sin^{4} (5 x) d x

2 y y^{^{'}} + 6 = y^{2} + 6 x, y (0) = 7

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