quick, whats the antiderivative of (sinx)^2

cirthix

Diamond Member
Aug 28, 2004
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please
dumb me, i forgot to look at double angle formulas, problem solved, thanks guys
all done :)
 

cirthix

Diamond Member
Aug 28, 2004
3,616
1
76
i'm doing homework, skipped that problem, but i'll be done with everything but that problem in like 10 minutes, an dwould like to have it all done
 

cirthix

Diamond Member
Aug 28, 2004
3,616
1
76
cool link, but i can't get it to work properly, it keeps thinking s, i, and n are variables lol

edit: brackets not parentheses, bleh. awesome link though, so fun!
 

cmp1223

Senior member
Jun 7, 2004
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tricky... maybe think of it as (sinx)(sinx) and use integration by parts

u=sinx dv=sinx
du=cosx v=-cosx

= (sinx)(-cosx)- integral[-cosx(cos)}

yikes.. it starts over :( sorry
 

Darien

Platinum Member
Feb 27, 2002
2,817
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You have to use a double angle formula for the integration:

Cos[2x] = 1 - 2*Sin[x]*Sin[x]