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$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}}=$
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The correct answer is:
$7 \pi / 44$
$I=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{d x}{1+\sqrt{\tan x}}$
$I=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$...(i)
$=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}}{\sqrt{\sin \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}+\sqrt{\cos \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}} d x$
$I=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$...(ii)
Adding (i) and (ii)
$\begin{aligned} & =\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} 1 \mathrm{dx}=[\mathrm{x}]_{\pi / 11}^{9 \pi / 22} \\ & =\frac{7 \pi}{22} \Rightarrow \mathrm{I}=\frac{7 \pi}{44}\end{aligned}$
$I=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$...(i)
$=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}}{\sqrt{\sin \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}+\sqrt{\cos \left(\frac{9 \pi}{22}+\frac{\pi}{11}-x\right)}} d x$
$I=\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$...(ii)
Adding (i) and (ii)
$\begin{aligned} & =\int_{\frac{\pi}{11}}^{\frac{9 \pi}{22}} 1 \mathrm{dx}=[\mathrm{x}]_{\pi / 11}^{9 \pi / 22} \\ & =\frac{7 \pi}{22} \Rightarrow \mathrm{I}=\frac{7 \pi}{44}\end{aligned}$
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