If int(sqrt{operatorname{cosec} x+1}) d x=k t | vedclass

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(A) Let $I = \int \sqrt{\operatorname{cosec} x + 1} dx = \int \sqrt{\frac{1}{\sin x} + 1} dx = \int \sqrt{\frac{1 + \sin x}{\sin x}} dx$.Using $\sin x

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$\frac{1}{2}$

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(A) Let $I = \int \sqrt{\operatorname{cosec} x + 1} dx = \int \sqrt{\frac{1}{\sin x} + 1} dx = \int \sqrt{\frac{1 + \sin x}{\sin x}} dx$.Using $\sin x = \frac{2 	an(x/2)}{1 + 	an^2(x/2)}$ and $1 + \sin x = (\sin(x/2) + \cos(x/2))^2$, we substitute $u = 	an(x/2)$, $dx = \frac{2}{1+u^2} du$.The integral becomes $I = \int \sqrt{\frac{(1+u)^2}{2u}} \cdot \frac{2}{1+u^2} du = \sqrt{2} \int \frac{1+u}{\sqrt{u}(1+u^2)} du$.Let $v = \sqrt{u}$, then $dv = \frac{1}{2\sqrt{u}} du$, so $du = 2v dv$.$I = 2\sqrt{2} \int \frac{1+v^2}{1+v^4} dv = \sqrt{2} \int \frac{1+1/v^2}{v^2+1/v^2} dv = \sqrt{2} \int \frac{d(v-1/v)}{(v-1/v)^2 + 2} + \sqrt{2} \int \frac{d(v+1/v)}{(v+1/v)^2 - 2}$.Evaluating this leads to $I = 2 	an^{-1}\left(\frac{\sqrt{2	an(x/2)}}{1-	an(x/2)}ight) + c$.Thus, $k = 2$ and $f(x) = \frac{\sqrt{2	an(x/2)}}{1-	an(x/2)}$.For $x = \pi/6$, $	an(x/2) = 	an(\pi/12) = 2 - \sqrt{3}$.$f(\pi/6) = \frac{\sqrt{2(2-\sqrt{3})}}{1-(2-\sqrt{3})} = \frac{\sqrt{4-2\sqrt{3}}}{\sqrt{3}-1} = \frac{\sqrt{(\sqrt{3}-1)^2}}{\sqrt{3}-1} = 1$.Therefore, $\frac{1}{k} f(\pi/6) = \frac{1}{2} \cdot 1 = \frac{1}{2}$.

If $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$, then $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$

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