What is the integral from 0 to pi/4 of 8sec^3(x) ?

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

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integral from 0 to pi/4 of 8sec^3(x)

\int_{0}^{\frac{π}{4}} 8 sec^{3} (x) d x

4 (ln (1 + 2) + 2)

Decimal Notation

9.18234 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 8 sec^{3} (x) d x

Take the constant out:

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

= 8 \cdot \int_{0}^{\frac{π}{4}} sec^{3} (x) d x

Apply Integral Reduction

= 8 ([\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)

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2}

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

Simplify

= 4 (ln (1 + 2) + 2)

\frac{t d Q}{d t} + Q = t^{4} ln (t)

\frac{d}{d x} (\frac{1}{e ^{x}})

f (x) = 2 x^{2} - 5 x + 1, at x = 2

\int \frac{x ^{3}}{4 - x ^{2}} d x

(- 3, 1), (2, - 1)

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