Let $I = \int (\sqrt{ an x} + \sqrt{\cot x}) dx$. Which of the following is/are mathematically equi

Question

Let $I = \int (\sqrt{	an x} + \sqrt{\cot x}) dx$. Which of the following is/are mathematically equivalent expressions for $I$? (where $C, C'$ are arbitrary constants of integration)

Professor PC Thomas Classes · Answer

Rewrite as $\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} = \sqrt{2} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}$. Let $t = \sin x - \cos x$, then $dt = (\cos x + \sin x)dx$. The integral becomes $\sqrt{2} \int \frac{dt}{\sqrt{1-t^2}} = \sqrt{2} \sin^{-1}(t) + C = \sqrt{2} \sin^{-1}(\sin x - \cos x) + C$ (Option A). Since $\sin^{-1}u = \pi/2 - \cos^{-1}u$, Option B is identical modulo the constant. Furthermore, if $\sin 	heta = u$, then $	an 	heta = u/\sqrt{1-u^2}$. Here $u = \sin x - \cos x$, so $1-u^2 = 2\sin x\cos x$. Hence Option C is also identical.