What is the integral from pi/2 to pi of (1+cos(x))/x ?

The integral from pi/2 to pi of (1+cos(x))/x is ln(2)+\Ci(pi)-\Ci(pi/2)

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integral from pi/2 to pi of (1+cos(x))/x

\int_{\frac{π}{2}}^{π} \frac{1 + cos ( x )}{x} d x

ln (2) + Ci (π) - Ci (\frac{π}{2})

Solution steps

\int_{\frac{π}{2}}^{π} \frac{1 + cos ( x )}{x} d x

Expand

\frac{1 + cos ( x )}{x} : \frac{1}{x} + \frac{cos ( x )}{x}

= \int_{\frac{π}{2}}^{π} \frac{1}{x} + \frac{cos ( 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

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

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

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

= ln (2) + Ci (π) - Ci (\frac{π}{2})

\int_{1}^{3} (\frac{1}{x} - y) d x

\int_{0}^{\frac{π}{2}} cos (9 x) cos (18 x) d x

\int_{1}^{3} x^{5} ln (x) d x

\int_{0}^{4} (4 x^{2} + x y - 2 y^{2}) d x

\int_{0}^{\frac{1}{2}} x sin (π x) d x

What is the integral from pi/2 to pi of (1+cos(x))/x ?
The integral from pi/2 to pi of (1+cos(x))/x is ln(2)+\Ci(pi)-\Ci(pi/2)