How do you integrate $tanx / (cosx)^2$ ?

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Answer 1 · 2016-10-18T16:55:40

$inttanx/(cosx)^2dx=1/2tan^2x+C$

Let $u=tanx$ , hence $du=sec^2xdx$

Hence $inttanx/(cosx)^2dx$

= $inttanxsec^2xdx$

= $intudu$

= $u^2/2$

= $1/2tan^2x+C$

Answer 2 · 2016-10-19T23:39:00

$sec^2x/2+C" "$ or $" "tan^2x/2+C$

$I=inttanx/cos^2xdx$

Since $tanx=sinx/cosx$ :

$I=intsinx/cos^3xdx$

We will use the substitution $u=cosx$ , which implies that $du=-sinxdx$ :

$I=-int(-sinx)/cos^3xdx$

$I=-intu^-3du$

Using the typical power rule for integration:

$I=-(u^-2/(-2))+C$

$I=1/(2u^2)+C$

$I=1/(2cos^2x)+C$

$I=sec^2x/2+C$

This is equivalent to the answer found by Shwetank Mauria, as $sec^2x$ and $tan^2x$ are separated by a constant: $sec^2x=tan^2x+1$ .

$I=(tan^2x+1)/2+C$

$I=tan^2x/2+1/2+C$

The $1/2$ is absorbed into $C$ ;

$I=tan^2x/2+C$

Both answers are valid.

How do you integrate tanx / (cosx)^2tanx / (cosx)^2?