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\( \int \frac{\cos (2 x)-1}{\cos (2 x)+1} d x= \) |
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Answer» \(\int \frac{Cos2x-1}{cos2x+1}dx\) = \(\int \frac{1-2\,sin^2x-1}{2cos^2-1+1}dx\) = \(\int \frac{-2\,sin^2x}{2\,cos^2x}dx\) = \(-\int tan^2x\,dx\) = \(-\int (sec^2x-1)\,dx\) = \(-\int sec^2xdx+\int dx\) = \(x-tanx+c\) |
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