Solving a Trig Substitution Problem: Incorrect Solution

AI Thread Summary
The discussion revolves around finding an appropriate trigonometric substitution for the integral \(\int\frac{x}{\sqrt{-29-4x^2-24x}}\). The initial attempt involved completing the square and suggested using \(\sqrt{7/4}\sec(t)-3\), which was deemed incorrect. Participants noted that using \(x=\sqrt{7/4}\sin(t)-3\) does not work, but substituting with \(\cosh t\) is a viable alternative. The equation \(-29-4x^2-24x\) can be rewritten to facilitate substitution, but care must be taken to ensure the integrand remains real. The conversation highlights the importance of accurate substitutions in solving trigonometric integrals.
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can someone help me find a appropriate trig sub for this problem:

\int\frac{x}{sqrt(-29-4x^2-24x)}

took out sqrt(4)...

sqrt(4)*sqrt(-29/4-x^2-6x)
(i also changed all the negative signs to positive)
complete the square...

sqrt(4)*sqrt((x+3)^2-7/4)

so my trig sub should be sqrt(7/4)*sec(t)-3, but it's incorrect
 
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ProBasket said:
can someone help me find a appropriate trig sub for this problem:

\int\frac{x}{sqrt(-29-4x^2-24x)}


Is it \int\frac{xdx}{sqrt(-29-4x^2-24x)} or

\int\frac{dx}{sqrt(-29-4x^2-24x)}?

so my trig sub should be sqrt(7/4)*sec(t)-3, but it's incorrect



Try x=\sqrt{7/4}*\sin(t)-3

ehild
 
It doesn't work with "sin",but it works with \cosh t...

Daniel.
 
dextercioby said:
It doesn't work with "sin",but it works with \cosh t...

Daniel.


-29-4x^2-24x = 7-(2x+6)^2=7(1-(\frac{2x+6}{7})^2)
If (2x+6)/\sqrt{7} \le 1 then it can be substituted either by sin(t) or cos(t). In the opposite case, the integrand is imaginary and your substitution is OK.

ehild
 
I don't mean to be picky,but you left out a square root sign...:wink:

Daniel.
 
dextercioby said:
I don't mean to be picky,but you left out a square root sign...:wink:

Daniel.

Oppps, really, you are right... :blushing:

ehild
 

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