int(sqrt{tan x}+sqrt{cot x}) d x= | vedclass

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Question

(A) Let $I = \int(\sqrt{	an x} + \sqrt{\cot x}) dx$. We can write this as $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}

Vedclass · Accepted Answer

$\sqrt{2} \sin ^{-1}(\sin x-\cos x)+c$,where $c$ is a constant of integration.

Vedclass · Answer

(A) Let $I = \int(\sqrt{	an x} + \sqrt{\cot x}) dx$. We can write this as $I = \int \left(\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}ight) dx$. Simplifying the integrand,we get $I = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} dx$. Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$. Squaring both sides of $t = \sin x - \cos x$,we get $t^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - 2 \sin x \cos x$. Thus,$2 \sin x \cos x = 1 - t^2$,which implies $\sin x \cos x = \frac{1 - t^2}{2}$. Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{\frac{1 - t^2}{2}}} = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}}$. Integrating,we obtain $I = \sqrt{2} \sin^{-1}(t) + c$. Substituting $t = \sin x - \cos x$ back,we get $I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + c$.

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

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