int (sqrt{tan x} + sqrt{cot x}) , dx is equal | vedclass

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Question

(B) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. Rewrite the integrand: $\sqrt{	an x} + \sqrt{\cot x} = \sqrt{\frac{\sin x}{\cos x}} + \sqrt

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$\sqrt{2} \sin^{-1} (\sin x - \cos x) + C$

Vedclass · Answer

(B) Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) \, dx$. Rewrite the integrand: $\sqrt{	an x} + \sqrt{\cot x} = \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} = \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}$. Multiply numerator and denominator by $\sqrt{2}$: $\frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} = \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{\sin 2x}}$. Since $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin^2 x + \cos^2 x - 2 \sin x \cos x) = 1 - (\sin x - \cos x)^2$,we have: $I = \sqrt{2} \int \frac{(\sin x + \cos x) \, dx}{\sqrt{1 - (\sin x - \cos x)^2}}$. Let $u = \sin x - \cos x$,then $du = (\cos x + \sin x) \, dx$. Substituting this into the integral: $I = \sqrt{2} \int \frac{du}{\sqrt{1 - u^2}} = \sqrt{2} \sin^{-1}(u) + C$. Thus,$I = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$.

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$ is equal to-

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