How do you integrate #int 1/sqrt(4x^2+4x+-24)dx# using trigonometric substitution?

Question

Luke Alvarez · Accepted Answer

#I=ln|(2x+1+sqrt(4x^2+4x-24))/5|+C,#

OR

#I=ln|2x+1+sqrt(4x^2+4x-24)|+K,#
where, #K=C-ln5.# I will solve #I=int1/sqrt(4x^2+4x-24)dx.# We have, #4x^2+4x-24=4x^2+4x+1-25=(2x+1)^2-5^2.# Knowing that, #sec^2theta-1=tan^2theta, 2x+1=5sect# will be proper substitution. So, let, #2x+1=5sect............(1),# so that, #2dx=5*sect*tantdt,# and, #sqrt(4x^2+4x-24)=sqrt{(2x+1)^2-5^2}=sqrt(25sec^2t-25)=5tant...........(2).# Hence, #I=int1/sqrt(4x^2+4x-24)dx=int1/(5tant)*5*sect*tantdt=intsectdt=ln|sect+tant|.# Hence, by #(1) & (2), I=ln|(2x+1+sqrt(4x^2+4x-24))/5|+C.# One may say #I=ln|2x+1+sqrt(4x^2+4x-24)|+K,# where, #K=C-ln5.#

Emily Ammerman · Answer

undefined To integrate $ \int \frac{1}{\sqrt{4x^2 + 4x - 24}} \, dx $ using trigonometric substitution, we'll complete the square in the denominator and then apply the appropriate substitution.

First, we'll rewrite the expression inside the square root:

$$ 4x^2 + 4x - 24 = 4(x^2 + x - 6) $$

Now, we complete the square within the parentheses:

$$ x^2 + x - 6 = (x + \frac{1}{2})^2 - (\frac{1}{2})^2 - 6 $$
$$ = (x + \frac{1}{2})^2 - \frac{1}{4} - 6 $$
$$ = (x + \frac{1}{2})^2 - \frac{25}{4} $$

Now, we see that this expression resembles the form $ a^2 - b^2 $, which motivates the substitution $ x + \frac{1}{2} = \frac{5}{2}\sec(\theta) $, where $ \sec(\theta) = \frac{1}{\cos(\theta)} $.

Then, we find $ dx $ in terms of $ d\theta $:

$$ dx = \frac{5}{2}\sec(\theta)\tan(\theta) \, d\theta $$

Now, substitute $ x $ and $ dx $ into the integral:

$$ \int \frac{1}{\sqrt{4x^2 + 4x - 24}} \, dx = \int \frac{1}{\sqrt{25\sec^2(\theta) - 25}} \cdot \frac{5}{2}\sec(\theta)\tan(\theta) \, d\theta $$

Simplify the expression inside the integral:

$$ = \int \frac{1}{\sqrt{25(\sec^2(\theta) - 1)}} \cdot \frac{5}{2}\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int \frac{1}{\sqrt{25\tan^2(\theta)}} \cdot \frac{5}{2}\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int \frac{1}{5|\tan(\theta)|} \cdot \frac{5}{2}\sec(\theta)\tan(\theta) \, d\theta $$

$$ = \int \frac{1}{|\tan(\theta)|} \cdot \frac{1}{2}\sec(\theta) \, d\theta $$

Now, integrate with respect to $ \theta $, remembering to consider the absolute value of $ \tan(\theta) $ for different intervals of $ \theta $. Finally, revert back to the variable $ x $ after integrating.