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

Question

Ezra Adams · Accepted Answer

#ln|sqrt((x + 1)^2 - 1) + x + 1| + C#

Complete the square in the denominator (within the #sqrt#).

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

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

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

Let #u = x + 1#. Then #du = dx#.

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

Now use the substitution #u = sectheta#. Then #du = secthetatanthetad theta#.

#int 1/sqrt(sec^2theta - 1) secthetatantheta d theta#

#int 1/sqrt(tan^2theta) secthetatantheta d theta#

#int 1/tantheta secthetatantheta d theta#

#int sectheta d theta#

This is a known integral that can be derived here

#ln|sectheta + tantheta| + C#

Obviously it's not good enough to stay in #theta#; we have to return to #x#. From our initial substitution, #u/1 = sectheta#. This means that the side opposite #theta# measures #sqrt(u^2 - 1)#. This also means that #tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)#.

#ln|sqrt(u^2 - 1) + u| + C#

We have one more substitution to reverse.

#ln|sqrt((x + 1)^2 - 1) + x + 1| + C#

Hopefully this helps!

Charles Arocha · Answer

undefined To integrate ∫(1/√(x^2 + 2x)) dx using trigonometric substitution, we follow these steps:

1. Complete the square: Rewrite the denominator in the form (x + 1)^2 - 1.
   x^2 + 2x = (x^2 + 2x + 1) - 1 = (x + 1)^2 - 1.

2. Apply the trigonometric substitution:
   Let x + 1 = √2 * tan(θ),
   then dx = √2 * sec^2(θ) dθ.

3. Substitute x + 1 = √2 * tan(θ) into the integral and express everything in terms of θ.

4. Solve for dx in terms of dθ.

5. Substitute dx in terms of dθ and simplify the integral.

6. Integrate the resulting expression with respect to θ.

7. Finally, express the result in terms of x by reversing the substitution used in step 2.

By following these steps, you can integrate ∫(1/√(x^2 + 2x)) dx using trigonometric substitution.