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JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If the integral \(\int {\frac{{\cos \,8x + 1}}{{\cot \,2x - \tan \,2x}}} dx = A\,\cos \,8x + k,\) where \(k\) is an arbitrary constant, then \(A\) is equal to
- A \( - \frac{1}{{16}}\)
- B \( \frac{1}{{16}}\)
- C \( \frac{1}{8}\)
- D \( - \frac{1}{8}\)
Answer & Solution
Correct Answer
(A) \( - \frac{1}{{16}}\)
Step-by-step Solution
Detailed explanation
Let \(1=\int \frac{\cos 8 x+1}{\cot 2 x-\tan 2 x} d x\) Now. \({{\rm{D}}^\prime } = \cot 2x - \tan 2x\) \( = \frac{{\cos 2x}}{{\sin 2x}} - \frac{{\sin 2x}}{{\cos 2x}}\) \( = \frac{{{{\cos }^2}2x - {{\sin }^2}2x}}{{\sin 2x\cos 2x}}\) \({ = \frac{{2\cos 4x}}{{\sin 4x}}}\)…
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