Q81 / 150 · MHT CET 2023 (09 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $\sqrt{2} \sin ^{-1}(\sin x-\cos x)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • B
    $\frac{1}{\sqrt{2}} \sin ^{-1}(\sin x-\cos x)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • C
    $\sin ^{-1}(\sin x-\cos x)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  • D
    $2 \sin ^{-1}(\sin x-\cos x)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int(\sqrt{\tan x}+\sqrt{\cot x}) \mathrm{d} x \\ & =\int\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right) d x \\ & \mathrm{I}=\int \frac{\sin x+\cos x}{\sqrt{\sin x \cos x}} \mathrm{~d} x \\ & \text { Let } \sin x-\cos x=\mathrm{t} \\ & \therefore \quad(\sin x+\cos x) \mathrm{d} x=\mathrm{dt} \\ & \text { Consider, } \sin x-\cos x=\mathrm{t} \\ & \text { Squaring on both sides, we get } \\ & 1-2 \sin t \cdot \cos t=t^2 \\ & 1-t^2=2 \sin t \cdot \cos t \\ & \therefore \quad \sin t \cdot \cos t=\frac{1-t^2}{2} \\ & \therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\frac{\sqrt{1-\mathrm{t}^2}}{\sqrt{2}}} \\ & \therefore \quad \mathrm{I}=\sqrt{2} \int \frac{1}{\sqrt{1-\mathrm{t}^2}} \mathrm{dt} \\ & \therefore \quad \mathrm{I}=\sqrt{2} \sin ^{-1}(\mathrm{t})+\mathrm{c} \\ & \therefore \quad \mathrm{I}=\sqrt{2} \sin ^{-1}(\sin x-\cos x)+\mathrm{c} \\ & \end{aligned}$

Hence, the correct answer is option: 'A'.

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