What is the integral from pi/6 to pi/4 of cot^2(x) ?

The integral from pi/6 to pi/4 of cot^2(x) is -pi/(12)-1+sqrt(3)

integral from pi/6 to pi/4 of cot^2(x)

\int_{\frac{π}{6}}^{\frac{π}{4}} cot^{2} (x) d x

- \frac{π}{12} - 1 + 3

Decimal

0.47025 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{4}} cot^{2} (x) d x

Rewrite using trig identities

= \int_{\frac{π}{6}}^{\frac{π}{4}} - 1 + csc^{2} (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}{6}}^{\frac{π}{4}} 1 d x + \int_{\frac{π}{6}}^{\frac{π}{4}} csc^{2} (x) d x

\int_{\frac{π}{6}}^{\frac{π}{4}} 1 d x = \frac{π}{12}

\int_{\frac{π}{6}}^{\frac{π}{4}} csc^{2} (x) d x = - 1 + 3

= - \frac{π}{12} - 1 + 3

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What is the integral from pi/6 to pi/4 of cot^2(x) ?
The integral from pi/6 to pi/4 of cot^2(x) is -pi/(12)-1+sqrt(3)