How do you integrate #int 1/sqrt(4x^2-12x+10) # using trigonometric substitution?

Question

Cameron Alexander · Accepted Answer

#=> 1/2 sinh^(-1) (2x-3 ) + c #

The first thing to do is rewrite #4x^2-12x+10# :

#4x^2-12x+10 -= 4(x-3/2)^2 + 1 #

#=> int dx/sqrt(4(x-3/2)^2 +1 ) #

Making the substitution:

#2(x-3/2) = sinhtheta #

#2dx = cosh theta d theta #

#4(x-3/2)^2 = sinh^2 theta #

#=> int (1/2 coshtheta d theta ) / sqrt(sinh^2theta +1 ) #

We know #cosh^2theta - sinh^2 theta = 1 => 1 + sinh^2 = cosh^2 theta #

#=> 1/2 int ( cosh theta d theta ) / cosh theta #

#=> 1/2int 1 d theta #

#=> 1/2 theta + c #

#2x- 3 = sinhtheta => theta = sinh^(-1)(2x-3) #

#=> 1/2 sinh^(-1) (2x-3 ) + c #

Ezra Adams · Answer

undefined To integrate $\int \frac{1}{\sqrt{4x^2 - 12x + 10}}$ using trigonometric substitution, follow these steps:

1. Complete the square inside the square root.
2. Substitute $x = \frac{5}{2} + \frac{1}{2}\cos(\theta)$.
3. Express $dx$ in terms of $d\theta$.
4. Substitute $x$, $dx$, and rewrite the expression in terms of $\theta$.
5. Use trigonometric identities to simplify the expression.
6. Integrate the simplified expression.
7. Finally, convert back to the original variable.

By following these steps, you can integrate the given expression using trigonometric substitution.