TS EAMCET · Maths · Indefinite Integration
\(\int(\cot x \cot (x+\alpha)+1) d x=\)
- 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 \(\tan \alpha \log \left(\mid \frac{\cos x}{|\sin (x+\alpha)|}\right)+c\)
Answer & Solution
Correct Answer
(A) \(\cot \alpha \log \left(\mid \frac{\sin x}{|\sin (x+\alpha)|}\right)+c\)
Step-by-step Solution
Detailed explanation
\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…
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