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Find the integrals of the functions tan4x |
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Answer» ∫tan4xdx = ∫(sec2x - 1)tan2x.dx = ∫sec2x.tan2x - ∫tan2x.dx = ∫(tanx)2sec2x - ∫(sec2x - 1)dx \(= \frac{tan^3x}{3} - tanx + x + c\) |
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