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

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

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

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

4 (- \frac{π}{3} + 3)

Decimal Notation

2.73941 \dots

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

= 4 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} - 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

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

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d x = \frac{π}{3}

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

= 4 (- \frac{π}{3} + 3)

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