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

1. ZeHanz

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:$\frac{ 1 }{ 81 } \int\limits_{}^{}\frac{ p-49 }{ \sqrt{p} }dp$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...

2. ZeHanz

Oops, forgot to compensate with ½ for the 2x, so replace 1/81 by 1/162, to get even!

3. ZeHanz

If that second step is a little hard to follow, then here's extra explanation: p=9u+49, so u= (p-49)/9 = 1/9 *(p-49). Also: dp=9du, so du = 1/9 * dp. That accounts for the 1/81. Extra factor 1/2 I forgot, gives you 1/162.

4. ZeHanz

@MajikDUSTY: I think it's your turn now ;)

5. Hoa

perfect way and new to me. thanks for that

6. Hoa

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

7. ZeHanz

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...

8. Hoa

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

9. ZeHanz

You're right. Glad to be of help.