What is the integral from 0 to pi/2 of 5sec^4(t/2) ?

The integral from 0 to pi/2 of 5sec^4(t/2) is 40/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi/2 of 5sec^4(t/2)

\int_{0}^{\frac{π}{2}} 5 sec^{4} (\frac{t}{2}) d t

\frac{40}{3}

Decimal Notation

13.33333 \dots

Solution steps

\int_{0}^{\frac{π}{2}} 5 sec^{4} (\frac{t}{2}) d t

Take the constant out:

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

= 5 \cdot \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

= 5 \cdot \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

= 5 \cdot 2 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) d u

Rewrite using trig identities

= 5 \cdot 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

= 5 \cdot 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

= 5 \cdot 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}

= 5 \cdot 2 (1 + \frac{1}{3})

Simplify

5 \cdot 2 (1 + \frac{1}{3}) : \frac{40}{3}

= \frac{40}{3}

\int \frac{- 1200 x}{25 - x ^{2}} d x

3.23 e^{- 61.4 t}

\int (\frac{10}{x}) d x

x \to 3 lim (\frac{( 4 - x )}{x - 3})

\frac{\partial}{\partial y} (x^{4} + y^{4})

What is the integral from 0 to pi/2 of 5sec^4(t/2) ?
The integral from 0 to pi/2 of 5sec^4(t/2) is 40/3