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

Question

1813 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T18:48:23

#=> 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 #