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

The integral from 0 to 1 of (sin(x))/(1-x) is -(-(-0.51117…+sin(1)(\Ci(1)-\Ci(0))))

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

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

- (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0))))

Solution steps

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

Apply u-substitution

: \int_{1}^{0} - \frac{sin ( - u + 1 )}{u} d u

= \int_{1}^{0} - \frac{sin ( - u + 1 )}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{sin ( - u + 1 )}{u} d u

Take the constant out:

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

= - (- \int_{0}^{1} \frac{sin ( - u + 1 )}{u} d u)

Rewrite using trig identities

= - (- \int_{0}^{1} \frac{- cos ( 1 ) sin ( u ) + cos ( u ) sin ( 1 )}{u} d u)

Expand

\frac{- cos ( 1 ) sin ( u ) + cos ( u ) sin ( 1 )}{u} : - \frac{cos ( 1 ) sin ( u )}{u} + \frac{sin ( 1 ) cos ( u )}{u}

= - (- \int_{0}^{1} - \frac{cos ( 1 ) sin ( u )}{u} + \frac{sin ( 1 ) cos ( u )}{u} d u)

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}^{1} \frac{cos ( 1 ) sin ( u )}{u} d u + \int_{0}^{1} \frac{sin ( 1 ) cos ( u )}{u} d u))

\int_{0}^{1} \frac{cos ( 1 ) sin ( u )}{u} d u = 0.51117 \dots

\int_{0}^{1} \frac{sin ( 1 ) cos ( u )}{u} d u = sin (1) (Ci (1) - Ci (0))

= - (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0))))

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What is the integral from 0 to 1 of (sin(x))/(1-x) ?
The integral from 0 to 1 of (sin(x))/(1-x) is -(-(-0.51117…+sin(1)(\Ci(1)-\Ci(0))))