What is the integral from 0 to x of ((sin(t))/t-1) ?

The integral from 0 to x of ((sin(t))/t-1) is \Si(x)-x

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integral from 0 to x of ((sin(t))/t-1)

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

Si (x) - x

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} \frac{sin ( t )}{t} d t - \int_{0}^{x} 1 d t

\int_{0}^{x} \frac{sin ( t )}{t} d t = Si (x)

\int_{0}^{x} 1 d t = x

= Si (x) - x

What is the integral from 0 to x of ((sin(t))/t-1) ?
The integral from 0 to x of ((sin(t))/t-1) is \Si(x)-x