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How to show that this improper integral converges and how to compute its value? $$ I=\int_{0}^{\frac\pi 2}\frac{\cos(2t)}{\sqrt{\sin(2t)}}\mathrm{d}t. $$ I used that the integrated function is odd so it suffice to study the integral over $]0,\frac \pi 4]$. To prove that the last one converges, I used the change of variable $u=\sin(2t)$ which lead to the study of a Riemann integral that converges. So, it follows that the value of $I$ is $0$ because of the odd function. Is that the correct way to do it?

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    $\begingroup$ I think that neither the integrand nor the "primitive" are odd. $\endgroup$ – user3621272 Apr 19 '15 at 20:02
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$$\int_{0}^{\pi/2}\frac{\cos(2t)}{\sqrt{\sin(2t)}}\,dt = \frac{1}{2}\int_{0}^{\pi}\frac{\cos t}{\sqrt{\sin t}}\,dt =\frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{-\sin t}{\sqrt{\cos t}}\,dt=0$$ since we have an odd Riemann-integrable function integrated on a symmetric interval with respect to the origin. Riemann integrability follows from: $$\frac{d}{dt}\sqrt{\sin t}=\frac{\cos t}{2\sqrt{\sin t}}.$$

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Along the same lines as Jack's answer, we can use the substitution $t\mapsto\frac\pi2-t$ to get $$ \begin{align} &\int_0^{\pi/2}\frac{\cos(2t)}{\sqrt{\sin(2t)}}\,\mathrm{d}\tag{1}\\ &=\int_0^{\pi/2}\frac{\cos(\pi-2t)}{\sqrt{\sin(\pi-2t)}}\,\mathrm{d}t\tag{2}\\ &=-\int_0^{\pi/2}\frac{\cos(2t)}{\sqrt{\sin(2t)}}\,\mathrm{d}t\tag{3}\\[6pt] &=0\tag{4} \end{align} $$ where $(4)$ is the average of $(1)$ and $(3)$.

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