What is the integral from-h to η of cosh^2(k(h+z)) ?

The integral from-h to η of cosh^2(k(h+z)) is (η+h)/2+1/(4k)sinh(2k(η+h))

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integral from-h to η of cosh^2(k(h+z))

\int_{- h}^{η} cosh^{2} (k (h + z)) d z

\frac{η + h}{2} + \frac{1}{4 k} sinh (2 k (η + h))

Solution steps

\int_{- h}^{η} cosh^{2} (k (h + z)) d z

Apply u-substitution

: \int_{0}^{h + η} cosh^{2} (k u) d u

= \int_{0}^{h + η} cosh^{2} (k u) d u

Apply u-substitution

: \int_{0}^{k (h + η)} \frac{cosh ^{2} ( v )}{k} d v

= \int_{0}^{k (h + η)} \frac{cosh ^{2} ( v )}{k} d v

Take the constant out:

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

= \frac{1}{k} \cdot \int_{0}^{k (h + η)} cosh^{2} (v) d v

Rewrite using trig identities

= \frac{1}{k} \cdot \int_{0}^{k (h + η)} \frac{1 + cosh ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{k} \cdot \frac{1}{2} \cdot \int_{0}^{k (h + η)} 1 + cosh (2 v) 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

= \frac{1}{k} \cdot \frac{1}{2} (\int_{0}^{k (h + η)} 1 d v + \int_{0}^{k (h + η)} cosh (2 v) d v)

\int_{0}^{k (h + η)} 1 d v = k (η + h)

\int_{0}^{k (h + η)} cosh (2 v) d v = \frac{1}{2} sinh (2 k (η + h))

= \frac{1}{k} \cdot \frac{1}{2} (k (η + h) + \frac{1}{2} sinh (2 k (η + h)))

Simplify

\frac{1}{k} \cdot \frac{1}{2} (k (η + h) + \frac{1}{2} sinh (2 k (η + h))) : \frac{η + h}{2} + \frac{1}{4 k} sinh (2 k (η + h))

= \frac{η + h}{2} + \frac{1}{4 k} sinh (2 k (η + h))

\int_{0}^{1} \frac{x ^{2}}{4 x ^{3} + 1} d x

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\int_{6}^{7} 1

\int_{- 1}^{1} x^{3} - \frac{3}{5} x d x

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

What is the integral from-h to η of cosh^2(k(h+z)) ?
The integral from-h to η of cosh^2(k(h+z)) is (η+h)/2+1/(4k)sinh(2k(η+h))