## anonymous 5 years ago integrate 1/((x^5)*sqrt(9*x^2-1))

1. anonymous

I've gotten pretty far I think, set x= 1/3 sec(U)

2. anonymous

then I got to the point where I had $81 \int\limits_{}^{} (\cos(x))^4 dx$

3. anonymous

the I apply the reduction formula twice and sub back in values for x to get (81/4)*(1/(3*x))^(3)*((sqrt(9 x^2-1))/(3*x))+243/(8)*(1/(3*x))*((sqrt(9 x^2-1))/(3*x))+243

4. anonymous

|dw:1327038086064:dw|

5. anonymous

maple 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.

6. anonymous

I 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^2-1))) for some reason.

7. Mr.Math

It seems like if you have made a mistake when substituting back for x.

8. anonymous

I did this |dw:1327038548902:dw|

9. anonymous

I know I've done it correctly right up until the substitution

10. anonymous

|dw:1327038727194:dw| then all I did was substitute the values for sin and cos I found from the triangle

11. anonymous

oops that last term should be over 8

12. Mr.Math

There's more than a way to substitute back, seems to me like wolfarm did this $\tan{u}=\sqrt{9x^2-1} \implies u=\tan^{-1}{\sqrt{9x^2-1}}.$

13. Mr.Math

There should be a term in the form $$ku$$ in your integral, k is a constant.

14. anonymous

Back 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?

15. anonymous

is the integral |dw:1327039058333:dw|

16. anonymous

I think that's where my missing term might be

17. Mr.Math

Yeah.

18. anonymous

ok because I had it set to 1. Then I will do the tan substitution you mentioned before and it should work out

19. Mr.Math

And this is the term where the arctan came from, because $$u=\tan^{-1}{\sqrt{9x^2-1}}$$.

20. Mr.Math

Exactly.

21. anonymous

Yeah, thanks a lot for your help!

22. Mr.Math

You're welcome.

23. anonymous

If you guys are still online I ran into problems with how this was working out.

24. anonymous

I somehow need u= -arctan(1/sqrt(9x^2−1))

25. Mr.Math

Why would you need that?

26. anonymous

|dw:1327040345712:dw|

27. anonymous

that is what it's telling me I'm missing from the end

28. anonymous

|dw:1327040463342:dw|

29. anonymous

any 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