Q111 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics

If $\int \frac{1}{1-\cos x} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c$, then one of the values of $\frac{\pi \alpha}{4}-\beta$ is

  • A
    $-\frac{\pi}{2}$
  • Correct answer
    B
    $\pi$
  • C
    0
  • D
    $\frac{\pi}{4}$

Solution

$\begin{aligned} & \text { } \int \frac{1}{1-\cos x} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & \Rightarrow \int \frac{1}{2 \sin ^2 \frac{x}{2}} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & \Rightarrow \frac{1}{2} \int \operatorname{cosec}^2 \frac{x}{2} d x=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & -\cot \frac{x}{2}+c=\tan \left(\frac{x}{\alpha}+\beta\right)+c \\ & \tan \left(\frac{x}{2}-\frac{\pi}{2}\right)=\tan \left(\frac{x}{\alpha}+\beta\right) \\ & \Rightarrow \alpha=2, \beta=\frac{-\pi}{2} \therefore \frac{\pi \alpha}{4}-\beta=\frac{\pi}{2}+\frac{\pi}{2}=\pi\end{aligned}$

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

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