What is the integral from 0 to 0.5 of xcosh(x^2+1) ?

The integral from 0 to 0.5 of xcosh(x^2+1) is 0.21335…

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integral from 0 to 0.5 of xcosh(x^2+1)

\int_{0}^{0.5} x cosh (x^{2} + 1) d x

0.21335 \dots

Solution steps

\int_{0}^{0.5} x cosh (x^{2} + 1) d x

Apply u-substitution

: \int_{1}^{1.25} \frac{cosh ( u )}{2} d u

= \int_{1}^{1.25} \frac{cosh ( u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{1.25} cosh (u) d u

Use the common integral:

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

= \frac{1}{2} [sinh (u)]_{1}^{1.25}

Simplify

= 0.5 [sinh (u)]_{1}^{1.25}

Compute the boundaries:

0.42671 \dots

= 0.5 \cdot 0.42671 \dots

Simplify

= 0.21335 \dots

\int_{0}^{π} x sin^{2} (x) d x

\int_{0}^{5} \frac{1}{125 - x ^{3}} d x

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

\int_{0}^{3} 6 d x

\int_{- 2}^{6} x d x

What is the integral from 0 to 0.5 of xcosh(x^2+1) ?
The integral from 0 to 0.5 of xcosh(x^2+1) is 0.21335…