# trig intergrationWatch

#1
I am having trouble doing the following:

∫(sec²xtan²x)dx

I dont really need a solution, so if your not botherd just tell me what to do eg if needed what subsitution to make. But if you want to do it for me feel free

thanks
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13 years ago
#2
this question was on the cambridge college test.
note that sec²x is the differential of tanx

∫(sec²xtan²x)dx = (tan3x)/3 +C
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13 years ago
#3
As Undry1 points out, this is a case of which you can find has the general solution .

This is why it's very useful to know the derivatives of common functions
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#4
Oh I see what you two mean, just did it.
Thanks for help
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13 years ago
#5
I suppose you could just 'know' that it did that.

But it works like this:

let u = tan x; du = sec^2 x dx;

that means for
∫ sec^2 x tan ^2 x dx., you get:

∫ sec^2 x u^2 (du/sec^2)

∫ u^2 du. = (u^3)/3 + C = (tan^3 x)/3 + C

(Sorry, if you already used that method lol)
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