How do I evaluate $int(sec^2(x))/(tan^2(x)) d x$ ?

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Answer 1 · 2015-03-04T19:10:53

The answer is: $-1/tanx+c$ .

$intsec^2x/tan^2xdx=int1/cos^2x*tan^(-2)xdx=$

$=tan^(-2+1)x/(-2+1)+c=-1/tanx+c$ .

This is because:

$int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c$ and the derivative of the function $tanx$ is $1/cos^2x.$

How do I evaluate int(sec^2(x))/(tan^2(x)) d xint(sec^2(x))/(tan^2(x)) d x?