## anonymous 5 years ago intergrate cos4x sin7x dx

1. anonymous

is it : $\cos^4xsin^7x?$

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

or wasn't that the question? ._.

9. anonymous

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

10. anonymous

although mine is the answer for his question :P

11. anonymous

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

12. anonymous

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

13. anonymous

:)

14. anonymous

^_^

15. anonymous

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

16. anonymous

still didn't take it

17. anonymous

Try www.aceyourcollegeclasses.com