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MajikDUSTY
Group Title
I really need help with this trig substitution please.
integral of (x^3)/sqrt(9(x^2)+49)
 one year ago
 one year ago
MajikDUSTY Group Title
I really need help with this trig substitution please. integral of (x^3)/sqrt(9(x^2)+49)
 one year ago
 one year ago

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ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
You have to do two different substitutions... First, set u=x², so du=2xdx. The integral becomes:\[\int\limits_{}^{}\frac{ udu }{ \sqrt{9u+49 }}\]Next, set p = 9u+49, so dp=9du, giving:\[\frac{ 1 }{ 81 } \int\limits_{}^{}\frac{ p49 }{ \sqrt{p} }dp\]Now you can split up the fraction:\[\frac{ 1 }{ 81 }\int\limits_{}^{}\left( p^{\frac{1}{2}}49p^{\frac{1}{2}} \right)dp\]This is not difficult anymore, I guess...
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
Oops, forgot to compensate with ½ for the 2x, so replace 1/81 by 1/162, to get even!
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
If that second step is a little hard to follow, then here's extra explanation: p=9u+49, so u= (p49)/9 = 1/9 *(p49). Also: dp=9du, so du = 1/9 * dp. That accounts for the 1/81. Extra factor 1/2 I forgot, gives you 1/162.
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
@MajikDUSTY: I think it's your turn now ;)
 one year ago

Hoa Group TitleBest ResponseYou've already chosen the best response.0
perfect way and new to me. thanks for that
 one year ago

Hoa Group TitleBest ResponseYou've already chosen the best response.0
Question: do we have to be back to u and then to x to get the final answer or just stop at p, since we substitute and respect to u and then to p dp ? just a little bit confuse
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
No, you have to go back to x. First go from p to u, then from u to x, so in the end you'll have a nice (ahem) formula with x...
 one year ago

Hoa Group TitleBest ResponseYou've already chosen the best response.0
got it. I'm not asker. I read it and recognize that the way you solve the problem is really new to me. Just confirm the stuff. Thanks a lot
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.1
You're right. Glad to be of help.
 one year ago
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