# Request for proof of the following identity.

How do we establish the following identity? \begin{align} &\int_{0}^{\frac{\pi}{2}} \frac{1}{(\sin(\theta) + \cos(\theta)) \sqrt{\sin(\theta) \cos(\theta)}} \,d{\theta} \\ \stackrel{\text{def}}{=} &\lim_{a \rightarrow 0^{+} \,,\, b \rightarrow (\frac{\pi}{2})^{-}} \int_{a}^{b} \frac{1}{(\sin(\theta) + \cos(\theta)) \sqrt{\sin(\theta) \cos(\theta)}} \,d{\theta} \\ = &\pi. \end{align}

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If we recall that $\sin(\theta)=\tan(\theta)\cos(\theta)$ then:
$$I=\int_{0}^{\frac{\pi}{2}} \frac{1}{(1 + \tan(\theta)) \sqrt{\tan(\theta)}} \frac{\,d{\theta}}{\cos^2(\theta)}$$
$$=\int_{0}^{\frac{\pi}{2}} \frac{1}{(1 + \tan(\theta)) \sqrt{\tan(\theta)}} \,d[\tan(\theta)]$$
$$=\int_{0}^{\infty} \frac{1}{(1 + z) \sqrt{z}} \,dz=\pi$$
 How did you determine the value of that last integral? – Kris Williams Oct 19 '12 at 4:35 As $\displaystyle \frac{d}{dx} [{\tan^{-1}}(\sqrt{x})] = \frac{1}{2(1 + x) \sqrt{x}}$, by the Fundamental Theorem of Calculus, we have $\displaystyle \int_{a}^{b} \frac{1}{(1 + x) \sqrt{x}} \,dx = 2[{\tan^{-1}}(\sqrt{b}) - {\tan^{-1}}(\sqrt{a})]$. – Haskell Curry Oct 19 '12 at 11:45