What is the integral from 0 to i of sinh(piz) ?

The integral from 0 to i of sinh(piz) is (e^{-pii}+e^{pii}-2)/(2pi)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to i of sinh(piz)

\int_{0}^{i} sinh (π z) d z

\frac{e ^{- πi} + e ^{πi} - 2}{2 π}

Solution steps

\int_{0}^{i} sinh (π z) d z

Rewrite using trig identities

= \int_{0}^{i} \frac{- e ^{- π z} + e ^{π z}}{2} d z

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{i} - e^{- π z} + e^{π z} d z

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}^{i} e^{- π z} d z + \int_{0}^{i} e^{π z} d z)

\int_{0}^{i} e^{- π z} d z = \frac{- e ^{- πi} + 1}{π}

\int_{0}^{i} e^{π z} d z = \frac{1}{π} (e^{πi} - 1)

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

Simplify

\frac{1}{2} (- \frac{- e ^{- πi} + 1}{π} + \frac{1}{π} (e^{πi} - 1)) : \frac{e ^{- πi} + e ^{πi} - 2}{2 π}

= \frac{e ^{- πi} + e ^{πi} - 2}{2 π}

\int_{0}^{2} 2 π y (y^{2}) d y

\int_{0}^{12} \frac{5}{1 + 0.2 t ^{2}} d t

\int_{0}^{\frac{π}{9}} tan (3 x) d x

\int_{0}^{2} 1 + 5 x^{2} d x

\int_{7}^{9} (x - 7)^{6} d x

What is the integral from 0 to i of sinh(piz) ?
The integral from 0 to i of sinh(piz) is (e^{-pii}+e^{pii}-2)/(2pi)