What is the integral from 0 to 1 of cosh^2(x) ?

The integral from 0 to 1 of cosh^2(x) is (4e^2+e^4-1)/(8e^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of cosh^2(x)

\int_{0}^{1} cosh^{2} (x) d x

\frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

Solution steps

\int_{0}^{1} cosh^{2} (x) d x

Rewrite using trig identities

= \int_{0}^{1} \frac{1 + cosh ( 2 x )}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{1} 1 + cosh (2 x) d x

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{0}^{1} 1 d x + \int_{0}^{1} cosh (2 x) d x)

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

\int_{0}^{1} cosh (2 x) d x = \frac{e ^{4} - 1}{4 e ^{2}}

= \frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}})

Simplify

\frac{1}{2} (1 + \frac{e ^{4} - 1}{4 e ^{2}}) : \frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

= \frac{4 e ^{2} + e ^{4} - 1}{8 e ^{2}}

\int_{1}^{9} 3 u d u

\int_{0}^{0.34} ln (1 + x^{2}) d x

\int_{- 3}^{0} x (x + 3) (x + 1) d x

\int_{0}^{10} 10 e^{0.02 t} d t

\int_{1}^{3} \frac{3}{5 x ^{2} + 5} d x

What is the integral from 0 to 1 of cosh^2(x) ?
The integral from 0 to 1 of cosh^2(x) is (4e^2+e^4-1)/(8e^2)