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lands91
Group Title
((sin^(5)x)/5)((Sin^(7)x)/7)+c need to make it come out to this solution. (1/70)sin5(x)(5cos(2x)+9)+c
 one year ago
 one year ago
lands91 Group Title
((sin^(5)x)/5)((Sin^(7)x)/7)+c need to make it come out to this solution. (1/70)sin5(x)(5cos(2x)+9)+c
 one year ago
 one year ago

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lands91 Group TitleBest ResponseYou've already chosen the best response.0
original integral \[\cos ^{3}xsin ^{4}x\] got the solution with some help of \[\frac{ \sin ^{5}x }{ 5 }\frac{ \sin ^{7}x }{ 7 } +c\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
ok, what you get after factoring out sin^5 x ?
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
(1cos2x)/2
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
thats about where I get lost
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
that is just sin^2 x \(\large \frac{ \sin ^{5}x }{ 5 }\frac{ \sin ^{7}x }{ 7 } = \sin^5 x[\frac{ 1 }{ 5 }\frac{ \sin ^{2}x }{ 7 }] \\ \huge =\sin^5 x[\frac{ 1 }{ 5 }\frac{ \dfrac{1\cos 2x}{2} }{ 7 }] \\ \huge \\ \huge =\sin^5 x[\frac{ 1 }{ 5 }\dfrac{1\cos 2x}{14} ] \)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
can you proceed ?
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
do u just multiple the sin5x back into it now
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
or do i set a common denominator of 70
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
set a common denominator of 70
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
which makes more sense
 one year ago

lands91 Group TitleBest ResponseYou've already chosen the best response.0
awesome. I see it now. was just over analysis of it. thank you
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i hope you'll get exactly what you want, welcome ^_^
 one year ago
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