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

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Answer 1 · 2015-03-04T19:10:53

The answer is: $- \frac{1}{tan x} + c$ $-1/tanx+c$ .

$\int \frac{{sec}^{2} x}{{tan}^{2} x} d x = \int \frac{1}{{cos}^{2} x} \cdot {tan}^{- 2} x d x =$ $intsec^2x/tan^2xdx=int1/cos^2x*tan^(-2)xdx=$

$= \frac{{tan}^{- 2 + 1} x}{- 2 + 1} + c = - \frac{1}{tan x} + c$ $=tan^(-2+1)x/(-2+1)+c=-1/tanx+c$ .

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 x∫sec2(x)tan2(x)dxint(sec^2(x))/(tan^2(x)) d x?