How do you find the antiderivative of $\frac{sec x}{1 + {tan}^{2} x}$ $secx / (1+tan^2x)$ ?

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How do you find the antiderivative of $\frac{sec x}{1 + {tan}^{2} x}$ $secx / (1+tan^2x)$ ?

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Answer 1 · 2015-04-13T15:21:56

Maybe with this kind of notation you will realize this integral is very simple : $\int \frac{sec (x)}{1 + {tan}^{2} (x)} d x = \frac{1}{cos (x)} \cdot \frac{1}{1 + {tan}^{2} (x)} d x$ $int(sec(x))/(1+tan^2(x))dx = 1/cos(x)*1/(1+tan^2(x))dx$

${(1 + {tan}^{2} (x))}^{- 1} = {cos}^{2} (x)$ $(1+tan^2(x))^(-1)=cos^2(x)$

Then $\int cos (x) d x$ $int cos(x) dx$

$= [sin (x)] + C$ $= [sin(x)]+C$

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How do you find the antiderivative of $\frac{sec x}{1 + {tan}^{2} x}$ $secx / (1+tan^2x)$ ?

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How do you find the antiderivative of secx / (1+tan^2x)secx1+tan2xsecx / (1+tan^2x)?

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How do you find the antiderivative of $\frac{sec x}{1 + {tan}^{2} x}$ $secx / (1+tan^2x)$ ?