Q67 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

$$ \int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)}= $$

  • Correct answer
    A
    $\tan \left(1+\log \left(\tan \frac{x}{2}\right)\right)+\mathrm{c}$, where $\mathrm{c}$ is constant of integration
  • B
    $\tan (1+\log (\tan x))+\mathrm{c}$, where $\mathrm{c}$ is constant of integration
  • C
    $\tan \left(\log \left(\tan \frac{x}{2}\right)\right)+\mathrm{c}$, where c is constant of integration.
  • D
    $\tan \left(\tan \frac{x}{2}\right)+c$, where c is constant of integration.

Solution

Let $\mathrm{I}=\int \frac{\operatorname{cosec} x \mathrm{~d} x}{\cos ^2\left(1+\log \tan \frac{x}{2}\right)} \mathrm{d} x$ Let $1+\log \left(\tan \frac{x}{2}\right)=\mathrm{t}$ Differentiating both sides w.r.t. t, we get $$ \begin{aligned} & \frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \times \frac{1}{2} \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \operatorname{cosec} x \mathrm{~d} x=\mathrm{dt} \\ & \therefore \quad \mathrm{I}=\int \frac{1}{\cos ^2 \mathrm{t}} \mathrm{dt}=\int \sec ^2 \mathrm{t} d \mathrm{t} \\ & =\tan (\mathrm{t})+\mathrm{c} \\ & =\tan \left(1+\log \left(\tan \frac{x}{2}\right)\right)+c \\ & \end{aligned} $$

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

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