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

The integral from 0 to 1 of sinh^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 sinh^2(x)

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

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

Decimal Notation

0.40671 \dots

Solution steps

\int_{0}^{1} sinh^{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_{0}^{\frac{π}{8}} e^{- x} sin (4 x) d x

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

\int_{1}^{2} \frac{15 x}{5 x ^{2} + 6} d x

\int_{0}^{4} 1 + 64 x^{2} d x

\int_{8}^{\infty} ln (x) d x

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