What is the integral from 0 to ln(2x) of sin(5e^t) ?

The integral from 0 to ln(2x) of sin(5e^t) is \Si(10x)-1.54993…

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integral from 0 to ln(2x) of sin(5e^t)

\int_{0}^{l n (2 x)} sin (5 e^{t}) d t

Si (10 x) - 1.54993 \dots

Solution steps

\int_{0}^{l n (2 x)} sin (5 e^{t}) d t

Apply u-substitution

: \int_{5}^{10 x} \frac{sin ( u )}{u} d u

= \int_{5}^{10 x} \frac{sin ( u )}{u} d u

Use the nonelementary integral:

\int \frac{sin ( x )}{x} d x = Si (x)

= [Si (u)]_{5}^{10 x}

Compute the boundaries:

Si (10 x) - 1.54993 \dots

= Si (10 x) - 1.54993 \dots

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

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

\int_{- 1}^{3} x^{2} + 2 x - 1 d x

\int_{\frac{π}{6}}^{\frac{π}{3}} cot^{3} (x) d x

\int_{0}^{π} t^{2} d t

What is the integral from 0 to ln(2x) of sin(5e^t) ?
The integral from 0 to ln(2x) of sin(5e^t) is \Si(10x)-1.54993…