How do you integrate #int dx/(4x^2+9)^2# using trig substitutions?

Question

Aurora Alvarado · Accepted Answer

#int dx/(4x^2+9)^2=1/108arctan((2x)/3)+x/(72x^2+162)+C# #int dx/(4x^2+9)^2# =#1/2int (2dx)/((2x)^2+3^2)^2# After using #2x=3tanu# and #2dx=3(secu)^2*du# transforms, this integral became #1/2int (3(secu)^2*du)/(81(secu)^4)# =#1/54int (cosu)^2*du# =#1/108int (1+cos2u)*du# =#1/108u+1/216sin2u+C# =#1/108u+1/216*(2tanu)/((tanu)^2+1)+C# After using #2x=3tanu#, #tanu=(2x)/3# and #u=arctan((2x)/3)# inverse transforms, I found #int dx/(4x^2+9)^2# =#1/108arctan((2x)/3)+1/216*(2*(2x)/3)/(((2x)/3)^2+1)+C# =#1/108arctan((2x)/3)+x/(72x^2+162)+C#

Charles Arocha · Answer

undefined To integrate $ \int \frac{dx}{(4x^2 + 9)^2} $ using trigonometric substitution, we follow these steps:

1. Recognize that the expression $4x^2 + 9$ resembles a squared trigonometric function.
2. Substitute $x = \frac{3}{2} \tan(\theta)$ to express the integral in terms of trigonometric functions.
3. Compute $dx$ using trigonometric identities.
4. Rewrite $4x^2 + 9$ in terms of trigonometric functions.
5. Perform the substitution and simplify the integral.
6. Integrate the simplified expression.
7. Substitute back the original variable.

Would you like a step-by-step explanation for each of these steps?