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

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

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

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

2 (ln (1 + 2) + 2)

Decimal

4.59117 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integral Reduction

= 4 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

= 4 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= 2 (ln (1 + 2) + 2)

x \to 4 lim (\frac{1}{( x - 4 ) ^{2}})

\int 10 - \frac{1}{16} x d x

\int \frac{1}{2 sin ^{2} ( \frac{x}{5} )} d x

x = ln (x y), at x = 1

2 x + 1

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