What is #int sec^2(x)/(sqrt(1-tan^2x)) dx#?

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Answer 1 · 2018-03-22T07:57:59

#sin^(-1)(tanx)+c#

#I=int(sec^2x/(sqrt(1-tan^2x)))dx#

us ethe substitution

#u=tanx#

#:. du=sec^2xdx#

#=>I=int((sec^2x)/sqrt(1-u^2))xx (du)/sec^2x#

#I=int (du)/sqrt(1-u^2)#

this is a standard integral

#I=sin^(-1)(u)+c#

#=sin^(-1)(tanx)+c#