Calculus integration (hard)

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December 13th 2012, 11:04 PM
asilvester635

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Calculus integration (hard)

Do I use integration by substitution?? this is a really tricky question for me
December 13th 2012, 11:35 PM
MarkFL

Re: Calculus integration (hard)

Yes, try the substitution:

$u=1-x\,\therefore\,du=-dx$

Then use:

$\sec(u)\tan(u)=\frac{d}{du}(sec(u))$
December 14th 2012, 09:06 AM
asilvester635

Re: Calculus integration (hard)

LOL so the final answer is sec(1-x)... wow it's that easy...
December 14th 2012, 11:07 PM
MarkFL

Re: Calculus integration (hard)

Not quite that easy...you need to correctly substitute for the differential to get:

$-\int\,d(\sec(u))=-\sec(u)+C=-\sec(1-x)+C$

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