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\int \tan ^{2}x\sec ^{4}x\,dx\,

The procedure with \tan x and \sec x is pretty much the same as with \sin x and \cos x. In both cases, we want to get to a place where we can do a substitution. With \sin x, a suitable du is \cos x. This is the difference.

{\frac  {d(\tan x)}{dx}}=\sec ^{2}x\,

{\frac  {d(\sec x)}{dx}}=\sec x\tan x\,

So, if you think about it, you can figure out the procedure here, or for any other problem of this type, yourself. When there is an even number of \sec x, turn all but 2 into \tan x and we'll have a whole bunch of \tan x and the derivative of \tan x and nothing else.

\int \tan ^{2}x\sec ^{4}x\,dx=\int \tan ^{2}x(\tan ^{2}x+1)\sec ^{2}x\,dx=\int (\tan ^{4}x+\tan ^{2}x)\sec ^{2}x\,dx\,

If we were to do a substitution, we'd do u=\tan x and du=\sec ^{2}xdx but we've done enough to not write out the steps.

\int (\tan ^{4}x+\tan ^{2}x)\sec ^{2}x\,dx={\frac  {1}{5}}\tan ^{5}x+{\frac  {1}{3}}\tan ^{3}x+C\,

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