Prove:

$$\displaystyle \huge \int_0^\frac{\pi}{2} \cot^{-1}{\sqrt{1+\csc{\theta}}} \, \text{d}\theta = \frac{\pi^2}{12}$$

Board index **‹** Computation of Integrals **‹** Inverse Cotangent of a Surd
## Inverse Cotangent of a Surd

Prove:

$$\displaystyle \huge \int_0^\frac{\pi}{2} \cot^{-1}{\sqrt{1+\csc{\theta}}} \, \text{d}\theta = \frac{\pi^2}{12}$$

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$$\displaystyle \huge \int_0^\frac{\pi}{2} \cot^{-1}{\sqrt{1+\csc{\theta}}} \, \text{d}\theta = \frac{\pi^2}{12}$$

1 post
• Page **1** of **1**