How do you integrate #int 1/sqrt(4x^2+12x-16) # using trigonometric substitution?

1 Answer
Jan 14, 2017

#1/2ln|(sqrt(4x^2 + 12x - 16) + 2x + 3)/5| + C#

Explanation:

Start by factoring the denominator.

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

#int1/(2sqrt(x^2 + 3x - 4))dx#

#1/2int1/sqrt(x^2 + 3x - 4)dx#

Now complete the square in the denominator.

#1/2int 1/sqrt(1(x^2 + 3x + 9/4 - 9/4) - 4)dx#

#1/2int 1/sqrt(1(x + 3/2)^2 - 9/4 - 4)dx#

#1/2int 1/sqrt((x + 3/2)^2 - 25/4)dx#

Now let #u = x + 3/2#: Then #du = dx#.

#1/2int 1/sqrt(u^2 - 25/4)du#

Apply the substitution #u = 5/2sectheta#. Then #du = 5/2secthetatantheta d theta#:

#1/2int 1/sqrt((5/2sectheta)^2 - 25/4)* 5/2secthetatantheta d theta#

#1/2int 1/sqrt(25/4sec^2theta - 25/4)* 5/2secthetatantheta d theta#

#1/2int 1/sqrt(25/4(sec^2theta - 1))* 5/2secthetatantheta d theta#

Apply #sec^2theta -1 = tan^2theta#:

#1/2int 1/sqrt(25/4tan^2theta)* 5/2secthetatantheta d theta#

#1/2int 1/(5/2tantheta)* 5/2secthetatantheta d theta#

#1/2int sec theta d theta#

This is a known integral: #intsecxdx = ln|secx + tanx|#. For more details on how to integrate this, go [here].(http://math2.org/math/integrals/more/sec.htm)

We must determine an expression for #sectheta# and #tantheta#. Draw a triangle:

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As you can see, #tantheta = sqrt(4u^2 - 25)/5# and #sectheta = (2u)/5#

#1/2ln|sqrt(4u^2 -25)/5 + (2u)/5| + C#

#1/2ln|(sqrt(4u^2 - 25) + 2u)/5| + C#

#1/2ln|(sqrt(4(x + 3/2)^2 - 25) + 2(x + 3/2))/5| + C#

#1/2ln|(sqrt(4x^2 + 12x - 16) + 2x + 3)/5| + C#

Hopefully this helps!