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

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

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

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

\frac{8}{3}

Decimal Notation

2.66666 \dots

Solution steps

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

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}

x \to 3 lim (\frac{x - 3}{x ^{2} - 9})

f (x) = (\frac{x + 7}{x + 9})^{6}

\int \frac{1}{4 x 16 x ^{2} - 25} d x

- 3 e^{- 3 x} + e^{4 x} \cdot 4

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