What is the integral from 0 to t of cosh(2u) ?

The integral from 0 to t of cosh(2u) is 1/2 sinh(2t)

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integral from 0 to t of cosh(2u)

\int_{0}^{t} cosh (2 u) d u

\frac{1}{2} sinh (2 t)

Solution steps

\int_{0}^{t} cosh (2 u) d u

Apply u-substitution

: \int_{0}^{2 t} cosh (v) \frac{1}{2} d v

= \int_{0}^{2 t} cosh (v) \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 t} cosh (v) d v

Use the common integral:

\int cosh (v) d v = sinh (v)

= \frac{1}{2} [sinh (v)]_{0}^{2 t}

Compute the boundaries:

sinh (2 t)

= \frac{1}{2} sinh (2 t)

\int_{- 2}^{2} \frac{32}{x ^{2} + 4} - x^{2} d x

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

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

\int_{0}^{2} [x] d x

\int_{0}^{2} (x^{2} - 5) d x

What is the integral from 0 to t of cosh(2u) ?
The integral from 0 to t of cosh(2u) is 1/2 sinh(2t)