What is the integral from 1 to x^9 of (sin(u))/u ?

The integral from 1 to x^9 of (sin(u))/u is \Si(x^9)-0.9460830704

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integral from 1 to x^9 of (sin(u))/u

\int_{1}^{x^{9}} \frac{sin ( u )}{u} d u

Si (x^{9}) - 0.9460830704

Solution steps

\int_{1}^{x^{9}} \frac{sin ( u )}{u} d u

Use the nonelementary integral:

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

= [Si (u)]_{1}^{x^{9}}

Compute the boundaries:

Si (x^{9}) - 0.9460830704

= Si (x^{9}) - 0.9460830704

\int_{0}^{2} - 4 x^{2} + 8 x^{3} d x

\int_{- 8}^{3} (- x^{2} - 3 x) d x

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

\int_{1}^{\infty} 2 x^{- 2} d x

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

What is the integral from 1 to x^9 of (sin(u))/u ?
The integral from 1 to x^9 of (sin(u))/u is \Si(x^9)-0.9460830704