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integral help
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Sure!
\[\int\limits_{}^{} \tan^2 x \sec^2 x dx\]
You need some trig trick and then u sub.
i'm tryi
ing to figure out the trig trick
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Remember that tan^2-1=sec^2
if u= tan x du =... ?
du= sec x ?
no, what you get after differentiating, tan x .. ? its pretty standard.
cos?
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if u=tan x, thn du= sec^2 x dx
ok, you need to revise all your differentiation formulas again.... d/dx (tan x) = sec^2 x
so, if u= tan x, du = sec^2x dx
\(\int\limits_{}^{} \tan^2 x (\sec^2 x dx) = \int u^2 du\) can you solve that ?
yeah. so tan x = u and sec x =du so it would be \[ \frac{ u ^{3} }{ 3 } + c\]
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then (tan ^3)/3 +c
right?
and resubstitute back u= tan x
yes, thats correct
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