What is the integral from pi/6 to pi/3 of cot^4(x) ?

The integral from pi/6 to pi/3 of cot^4(x) is (3sqrt(3)pi+16)/(18sqrt(3))

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integral from pi/6 to pi/3 of cot^4(x)

\int_{\frac{π}{6}}^{\frac{π}{3}} cot^{4} (x) d x

\frac{3 3 π + 16}{18 3}

Decimal Notation

1.03679 \dots

Solution steps

\int_{\frac{π}{6}}^{\frac{π}{3}} cot^{4} (x) d x

Apply Integral Reduction

= - [\frac{cot ^{3} ( x )}{3}]_{\frac{π}{6}}^{\frac{π}{3}} - \int_{\frac{π}{6}}^{\frac{π}{3}} cot^{2} (x) d x

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

= - [\frac{cot ^{3} ( x )}{3}]_{\frac{π}{6}}^{\frac{π}{3}} - (- \frac{π}{6} - \frac{1}{3} + 3)

Simplify

- [\frac{cot ^{3} ( x )}{3}]_{\frac{π}{6}}^{\frac{π}{3}} - (- \frac{π}{6} - \frac{1}{3} + 3) : - [\frac{1}{3} cot^{3} (x)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{π}{6} + \frac{1}{3} - 3

= - [\frac{1}{3} cot^{3} (x)]_{\frac{π}{6}}^{\frac{π}{3}} + \frac{π}{6} + \frac{1}{3} - 3

Compute the boundaries:

\frac{1}{9 3} - 3

= - (\frac{1}{9 3} - 3) + \frac{π}{6} + \frac{1}{3} - 3

Simplify

= \frac{3 3 π + 16}{18 3}

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What is the integral from pi/6 to pi/3 of cot^4(x) ?
The integral from pi/6 to pi/3 of cot^4(x) is (3sqrt(3)pi+16)/(18sqrt(3))