A. \(2 \cos (2 x+\pi)+k\) B. \(\log [\cos (2 x+\pi)]+k\) C. \(-\log [\cos (2 x+\pi)]+k \quad\) D. \(4 \cot (2 x+\pi)+k\)
Correct Answer: C
Explanation
\(\int 2 \tan (2 x+\pi) d x \) We bring out the constant \(=2 \int \tan (2 x+\pi) d x=2 \int \frac{\sin (2 x+\pi)}{\cos (2 x+\pi)} d x\) We apply double negative which do not alter the equation \(-\int \frac{-\sin (2 x+\pi)}{\cos (2 x+\pi)} d x\) but \(\int \frac{f^{\prime}(x)}{f(x)} d x=\log f(x)+c\) so that \(-\int \frac{-\sin (2 x+\pi)}{\cos (2 x+\pi)} d x=-\log [(\cos 2 x+\pi)+k\)