Evaluate the integral: int (sqrt{tan x} + sqr | vedclass

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Question

(B) 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}{\si

Vedclass · Accepted Answer

$\sqrt{2} 	an^{-1}\left(\frac{	an x - 1}{\sqrt{2 	an x}}ight) + C$

Vedclass · Answer

(B) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. We can write this as $I = \int \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} ight) \, dx = \int \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx$. Multiply numerator and denominator by $\sqrt{2}$: $I = \int \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} \, dx = \sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx$. Since $\sin 2x = 1 - (\sin x - \cos x)^2$,let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) \, dx$. Substituting these into the integral: $I = \sqrt{2} \int \frac{dt}{\sqrt{1 - t^2}} = \sqrt{2} \sin^{-1}(t) + C = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$. Using the identity $\sin^{-1}(u) = 	an^{-1}\left(\frac{u}{\sqrt{1-u^2}}ight)$,we get: $I = \sqrt{2} 	an^{-1}\left(\frac{\sin x - \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}ight) + C = \sqrt{2} 	an^{-1}\left(\frac{\sin x - \cos x}{\sqrt{\sin 2x}}ight) + C$. Dividing numerator and denominator by $\sqrt{\cos x}$,we obtain: $I = \sqrt{2} 	an^{-1}\left(\frac{	an x - 1}{\sqrt{2 	an x}}ight) + C$.

Evaluate the integral: $\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$

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