How do you integrate #int 1/(x^2sqrt(16x^2-9))# by trigonometric substitution?

Question

Aurora Alvarado · Accepted Answer

# sqrt(16x^2-9)/(9x)+C#.

Suppose that, #I=int1/(x^2sqrt(16x^2-9))dx=int1/(x^2sqrt{(4x)^2-3^2})dx#. We subst. #4x=3secy," so that, "4dx=3secytanydy#. #:. I=int1/{(3/4*secy)^2sqrt(9sec^2y-9)}*(3/4*secytany)dy#, #=16/9*1/4int1/(sec^2ytany)*secytanydy#, #=4/9int1/secydy#, #=4/9intcosydy#, #=4/9siny#, #=4/9sqrt(1-cos^2y)#, #=4/9sqrt(1-1/sec^2y)#, #=4/9sqrt{1-1/(4/3*x)^2}#, #=4/9sqrt{1-9/(16x^2)}#. # rArr I=sqrt(16x^2-9)/(9x)+C#.

Evelyn Agard · Answer

undefined To integrate $ \frac{1}{x^2\sqrt{16x^2 - 9}} $ using trigonometric substitution, perform the following steps:

1. Recognize the form $ \sqrt{a^2 - x^2} $, which suggests a trigonometric substitution.
2. Let $ x = \frac{3}{4}\sec(\theta) $ or $ x = \frac{3}{4}\csc(\theta) $, depending on whether $ a^2 - x^2 $ is in the form $ \sqrt{a^2 - x^2} $ or $ \sqrt{x^2 - a^2} $, respectively.
3. Express $ dx $ in terms of $ d\theta $.
4. Substitute $ x $ and $ dx $ in terms of $ \theta $.
5. Simplify the integral and solve it in terms of $ \theta $.
6. Finally, convert back to the original variable $ x $.

Following these steps should lead to the solution of the integral.