Answer to Question #108780 in Calculus for Nimra

Question #108780

∫tan x dx


1
Expert's answer
2020-04-13T14:21:49-0400

tanxdx=sinxcosxdxlett=cosxdt=sinxdxsinxcosxdx=dtt=lnt+c=lncosx+c\int\tan xdx=\int\frac{\sin x}{\cos x}dx\\ let\\ t=\cos x\\ dt=-\sin x dx\\ \int\frac{\sin x}{\cos x}dx=\int\frac{-dt}{t}=- \ln|t|+c=-\ln|\cos x|+c

tanxdx=lncosx+c\int\tan xdx=-\ln|cos x|+c


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