Question

intergrate cos4x sin7x dx

Answer 1

is it :

\[\cos^4xsin^7x?\]

Answer 2

1/2 (-1/11 cos11x -1/3cos3x_dx is the answer i got

Answer 3

-(1/6) Cos[3 x] - 1/22 Cos[11 x]

Answer 4

\[\int\limits_{}^{}\cos4xsin7x dx =1/2 \int\limits_{}^{}((\sin(3x)+\sin(11x))dx\]
(using that sin(7x)cos(4x)=1/2[sin(7x-4x)+sin(7x+4x)]

Answer 5

\[=-1/2[{1 \over 3}\cos(3x)+{1 \over 11} \cos(11x)] +c\]

Answer 6

\[{-1 \over 6} \cos(3x)-{1 \over 22} \cos(11x) +c\]

Answer 7

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

Answer 8

or wasn't that the question? ._.

Answer 9

he got two different answers for two different Questions.. lucky him :)

Answer 10

although mine is the answer for his question :P

Answer 11

lol, I'm not sure what was his question and considered it the following ^^"

Answer 12

oh then digger, ignore my answer lol, it's for a completely different question ^_^ follow anwar's steps

Answer 13

:)

Answer 14

^_^

Answer 15

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

Answer 16

still didn't take it

Answer 17

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