intergrate cos4x sin7x dx

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intergrate cos4x sin7x dx

Mathematics
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is it : \[\cos^4xsin^7x?\]
1/2 (-1/11 cos11x -1/3cos3x_dx is the answer i got
-(1/6) Cos[3 x] - 1/22 Cos[11 x]

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Other answers:

\[\int\limits_{}^{}\cos4xsin7x dx =1/2 \int\limits_{}^{}((\sin(3x)+\sin(11x))dx\] (using that sin(7x)cos(4x)=1/2[sin(7x-4x)+sin(7x+4x)]
\[=-1/2[{1 \over 3}\cos(3x)+{1 \over 11} \cos(11x)] +c\]
\[{-1 \over 6} \cos(3x)-{1 \over 22} \cos(11x) +c\]
if so then:\[\int\limits_{}^{}\cos^4x(1-\cos^2x)^4sinx dx\] take : u = cos x du = -sinx dx and you'll get : \[=-\int\limits_{}^{} u^4(1-u^2)^4 dx\] \[= -\int\limits_{}^{} u^4(u^8 -4u^6 + 2u^4 +1) du\] \[= -\int\limits_{}^{}(u^(12) -4u^(10) + 2u^8 +u^4) du\] \[= -[u^{13}/13 -4u^{11}/11 + 2u^{9}/9 + u^{5}/5] + c\] \[= -[\cos^{13}x/13 -4\cos^{11}x/11 +2\cos^{9}x/9 +\cos^{5}x/5] + c\] that's the answer if it were cos^4xsin^7x ^_^ correct me if I'm wrong
or wasn't that the question? ._.
he got two different answers for two different Questions.. lucky him :)
although mine is the answer for his question :P
lol, I'm not sure what was his question and considered it the following ^^"
oh then digger, ignore my answer lol, it's for a completely different question ^_^ follow anwar's steps
:)
^_^
no worry's. I'm sure AnwarA is correct as I got the same answer. What are you guys like with 2nd order linear differential equations with constantco-efficents
still didn't take it
Try www.aceyourcollegeclasses.com

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