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anonymous
 5 years ago
integrate 1/((x^5)*sqrt(9*x^21))
anonymous
 5 years ago
integrate 1/((x^5)*sqrt(9*x^21))

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I've gotten pretty far I think, set x= 1/3 sec(U)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then I got to the point where I had \[81 \int\limits_{}^{} (\cos(x))^4 dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the I apply the reduction formula twice and sub back in values for x to get (81/4)*(1/(3*x))^(3)*((sqrt(9 x^21))/(3*x))+243/(8)*(1/(3*x))*((sqrt(9 x^21))/(3*x))+243

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327038086064:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maple and wolfram tell me that there is supposed to be an arctan in there somehow. But I can't figure out where it would get added in.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can show more steps in between, but what I have there matches maple except for the last term 243/8 is multiplied by arctan(1/(sqrt(9 x^21))) for some reason.

Mr.Math
 5 years ago
Best ResponseYou've already chosen the best response.2It seems like if you have made a mistake when substituting back for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did this dw:1327038548902:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know I've done it correctly right up until the substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327038727194:dw then all I did was substitute the values for sin and cos I found from the triangle

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops that last term should be over 8

Mr.Math
 5 years ago
Best ResponseYou've already chosen the best response.2There's more than a way to substitute back, seems to me like wolfarm did this \[\tan{u}=\sqrt{9x^21} \implies u=\tan^{1}{\sqrt{9x^21}}.\]

Mr.Math
 5 years ago
Best ResponseYou've already chosen the best response.2There should be a term in the form \(ku\) in your integral, k is a constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Back in the step where you had \[81\int\limits_{}^{}\cos ^{4}(u) du\] did you consider substituting \[\cos^{2}u = (1 + \cos2u) / 2\] and solving from there?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the integral dw:1327039058333:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think that's where my missing term might be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok because I had it set to 1. Then I will do the tan substitution you mentioned before and it should work out

Mr.Math
 5 years ago
Best ResponseYou've already chosen the best response.2And this is the term where the arctan came from, because \(u=\tan^{1}{\sqrt{9x^21}}\).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, thanks a lot for your help!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you guys are still online I ran into problems with how this was working out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I somehow need u= arctan(1/sqrt(9x^2−1))

Mr.Math
 5 years ago
Best ResponseYou've already chosen the best response.2Why would you need that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327040345712:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is what it's telling me I'm missing from the end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dw:1327040463342:dw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0any ideas where this would come from. It's confusing since my first substitutions look right, but the u one isn't consistent with the same triangle
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