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∫ tan(x) dx =

∫ sin(x)/cos(x) dx

Let cos(x) = u

Therefore, sin(x) = -du/dx >>>> (Derivative of cos(x))

∫ (-du/dx)/u dx >>>>>>>>>>> Substitute the values

∫ -du/udx dx

∫ -du/u >>>>>>>>>>>>>>>>> dx * 1/dx cancel each other out

∫ -1/u du

= -ln(u) + C

= -ln(cos(x)) + C >>>>>>>>>>> Substitute cos(x) back into u

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Q: How do you find the integral of tanx?
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