Q114 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

$\int(\cot x \cot (x+\alpha)+1) d x=$

  • Correct answer
    A
    $\cot \alpha \log \left(\mid \frac{\sin x}{|\sin (x+\alpha)|}\right)+c$
  • B
    $\log |\sin x \sin (x+\alpha)|+x+c$
  • C
    $\log |\sin x \cos (x+\alpha)|+x+c$
  • D
    $\tan \alpha \log \left(\mid \frac{\cos x}{|\sin (x+\alpha)|}\right)+c$

Solution

$$ \begin{aligned} & \text { } \int[\cot x \cot (x+\alpha)+1] d x \\ & =\int \frac{\cos x \cos (x+\alpha)+\sin x \sin (x+\alpha)}{\sin x \sin (x+\alpha)} d x \\ & =\int \frac{\cos \alpha}{\sin x \sin (x+\alpha)} d x=\frac{\cos \alpha}{\sin \alpha} \int \frac{\sin \alpha}{\sin x \sin (x+\alpha)} d x \\ & =\cos \alpha \int \frac{\sin [(x+\alpha)-x]}{\sin x \sin (x+\alpha)} d x \\ & =\cot \alpha \int \frac{\sin (x+\alpha) \cos x-\cos (x+\alpha) \sin x}{\sin x \sin (x+\alpha)} d x \\ & =\cot \alpha \int(\cot x-\cot (x+\alpha)) d x \\ & =\cot \alpha[\log |\sin x|-\log |\sin (x+\alpha)|]+c \\ & =\cot \alpha \log \left|\frac{\sin x}{\sin (x+\alpha)}\right|+c \\ & \end{aligned} $$

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

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