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Evaluate: \(\int\frac{sin2x}{1 + cos^2x}dx\) |
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Answer» \(\int\frac{sin2x}{1 + cos^2x}dx\) = \(\int\frac{-dt}{t}\) = -logt + c = -log(1 + cos2x) + c put 1 + cos2x = t2 cosx(-sinx) dx = dt -sin2x dx = dt sin2xdx = -dt |
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