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

Question

Emma Aldridge · Accepted Answer

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

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].(https://tutor.hix.ai)

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

Cameron Alexander · Answer

undefined To integrate $ \int \frac{1}{\sqrt{4x^2 + 12x - 16}} $ using trigonometric substitution:

1. Complete the square inside the square root:
   $ 4x^2 + 12x - 16 = 4(x^2 + 3x - 4) = 4(x^2 + 3x + \frac{9}{4} - \frac{9}{4} - 4) = 4((x + \frac{3}{2})^2 - \frac{25}{4}) $

2. Recognize that $ (x + \frac{3}{2})^2 - \frac{25}{4} $ resembles the standard form of a trigonometric substitution, $ a^2 - b^2 = (a + b)(a - b) $.

3. Substitute $ x + \frac{3}{2} = \frac{5}{2} \sec(\theta) $, which gives $ dx = \frac{5}{2} \sec(\theta) \tan(\theta) d\theta $.

4. Rewrite the expression using the trigonometric substitution:
   $ \sqrt{4x^2 + 12x - 16} = \sqrt{4((x + \frac{3}{2})^2 - \frac{25}{4})} = \sqrt{4(\frac{25}{4} \sec^2(\theta))} = \sqrt{25 \sec^2(\theta)} = 5 \sec(\theta) $

5. Substitute $ dx $ and the expression under the square root into the integral, and rewrite it in terms of $ \theta $:
   $ \int \frac{1}{5 \sec(\theta)} \cdot \frac{5}{2} \sec(\theta) \tan(\theta) d\theta = \int 2 \tan(\theta) d\theta $

6. Integrate $ 2 \tan(\theta) $ with respect to $ \theta $:
   $ \int 2 \tan(\theta) d\theta = -2 \ln|\cos(\theta)| + C $

7. Substitute back for $ x $ using the original substitution, and simplify the result:
   $ -2 \ln|\cos(\theta)| + C = -2 \ln|\cos(\arccos(\frac{2}{5}(x + \frac{3}{2}))| + C = -2 \ln|\frac{2}{5}(x + \frac{3}{2})| + C $

So, $ \int \frac{1}{\sqrt{4x^2 + 12x - 16}} dx = -2 \ln|\frac{2}{5}(x + \frac{3}{2})| + C $.