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

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

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

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

4 - 4 2 + 2 - 22 2 + 2 + 7 + 42 + \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

Decimal Notation

0.80614 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{π}{16}} sec^{4} (u) \cdot 4 d u

= \int_{0}^{\frac{π}{16}} sec^{4} (u) \cdot 4 d u

Take the constant out:

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

= 4 \cdot \int_{0}^{\frac{π}{16}} sec^{4} (u) d u

Rewrite using trig identities

= 4 \cdot \int_{0}^{\frac{π}{16}} (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 + v^{2} d v

= 4 \cdot \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 d v + \int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} v^{2} d v

\int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} 1 d v = - 4 2 + 2 - 22 2 + 2 + 7 + 42

\int_{0}^{- 4 2 + 2 - 22 2 + 2 + 7 + 42} v^{2} d v = \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

= 4 - 4 2 + 2 - 22 2 + 2 + 7 + 42 + \frac{( - 4 2 + 2 - 2 2 2 + 2 + 7 + 4 2 ) ^{\frac{3}{2}}}{3}

\int_{0}^{\frac{x}{4}} e^{- x^{2}} d x

\int_{1}^{3} \frac{1}{x 4 + x ^{2}} d x

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

\int_{- 1}^{1} (2 - x^{2} - x^{2}) d x

\int_{0}^{2} \frac{1}{x ^{4}} d x

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