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

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

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

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

Si (- π + x) - Si (- π)

Solution steps

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

Factor

π - t : - (- π + t)

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

Take the constant out:

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

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

Apply u-substitution

: \int_{- π}^{- π + x} \frac{sin ( u + π )}{u} d u

= - \int_{- π}^{- π + x} \frac{sin ( u + π )}{u} d u

Rewrite using trig identities

= - \int_{- π}^{- π + x} - \frac{sin ( u )}{u} d u

Take the constant out:

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

= - (- \int_{- π}^{- π + x} \frac{sin ( u )}{u} d u)

Use the nonelementary integral:

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

= - (- [Si (u)]_{- π}^{- π + x})

Simplify

= [Si (u)]_{- π}^{- π + x}

Compute the boundaries:

Si (- π + x) - Si (- π)

= Si (- π + x) - Si (- π)

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