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

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

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 2 d u

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

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^{4} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int_{0}^{1} 1 + v^{2} d v

= 2 \cdot \int_{0}^{1} 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

= 2 (\int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v)

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} v^{2} d v = \frac{1}{3}

= 2 (1 + \frac{1}{3})

Simplify

2 (1 + \frac{1}{3}) : \frac{8}{3}

= \frac{8}{3}

\int_{- 4}^{4} x^{2} - 16 d x

\int_{0}^{1} \frac{1}{3 x - 2} d x

\int_{- 4}^{0} - y - (y^{2} + 3 y) d y

\int_{0}^{t} sin (u) d u

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

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