What is the integral from 0 to pi/4 of 6sec^3(x) ?

The integral from 0 to pi/4 of 6sec^3(x) is 3(ln(1+sqrt(2))+sqrt(2))

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integral from 0 to pi/4 of 6sec^3(x)

\int_{0}^{\frac{π}{4}} 6 sec^{3} (x) d x

3 (ln (1 + 2) + 2)

Decimal Notation

6.88676 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 6 sec^{3} (x) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (x) d x

Apply Integral Reduction

= 6 ([\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)

= 6 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Simplify

6 ([\frac{sec ^{2} ( x ) sin ( x )}{2}]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2)) : 6 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

= 6 ([\frac{1}{2} sec (x) tan (x)]_{0}^{\frac{π}{4}} + \frac{1}{2} ln (1 + 2))

Compute the boundaries:

\frac{1}{2}

= 6 (\frac{1}{2} + \frac{1}{2} ln (1 + 2))

Simplify

= 3 (ln (1 + 2) + 2)

x \to 2 - lim (\frac{( 2 ) ∣ x ∣}{x} - x)

\int 4 x - 1 d x

\int_{0}^{5} 6 x - x^{2} d x

\frac{d}{d x} (, x)

\frac{d y}{d x} = \frac{1}{x y ^{3}}

What is the integral from 0 to pi/4 of 6sec^3(x) ?
The integral from 0 to pi/4 of 6sec^3(x) is 3(ln(1+sqrt(2))+sqrt(2))