What is the integral from 1 to 2 of (sin^2(x))/x ?

The integral from 1 to 2 of (sin^2(x))/x is 1/2 (ln(2)-(\Ci(4)-\Ci(2)))

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integral from 1 to 2 of (sin^2(x))/x

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

\frac{1}{2} (ln (2) - (Ci (4) - Ci (2)))

Solution steps

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

Rewrite using trig identities

= \int_{1}^{2} \frac{1 - cos ( 2 x )}{2 x} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{2} \frac{1 - cos ( 2 x )}{x} d x

Expand

\frac{1 - cos ( 2 x )}{x} : \frac{1}{x} - \frac{cos ( 2 x )}{x}

= \frac{1}{2} \cdot \int_{1}^{2} \frac{1}{x} - \frac{cos ( 2 x )}{x} d x

Apply the Sum Rule:

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

= \frac{1}{2} (\int_{1}^{2} \frac{1}{x} d x - \int_{1}^{2} \frac{cos ( 2 x )}{x} d x)

\int_{1}^{2} \frac{1}{x} d x = ln (2)

\int_{1}^{2} \frac{cos ( 2 x )}{x} d x = Ci (4) - Ci (2)

= \frac{1}{2} (ln (2) - (Ci (4) - Ci (2)))

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What is the integral from 1 to 2 of (sin^2(x))/x ?
The integral from 1 to 2 of (sin^2(x))/x is 1/2 (ln(2)-(\Ci(4)-\Ci(2)))