Evaluate the indefinite integral:
INT dx/(64+x^2)^2
To evaluate the indefinite integral ∫ dx / (64 + x^2)^2, we can use a technique called substitution. Let's go step by step.
Step 1: First, let's identify a suitable substitution. We can make the substitution x = 8 tanθ. To understand why this substitution makes sense, let's differentiate it.
dx = 8 sec^2θ dθ
Now, we need to replace dx and x in terms of θ in the integral.
dx = 8 sec^2θ dθ
x = 8 tanθ
Step 2: Rewrite the integral in terms of θ.
To replace dx and x, we can substitute them in the integral:
∫ dx / (64 + x^2)^2 = ∫ (8 sec^2θ dθ) / (64 + (8 tanθ)^2)^2
Simplify the denominator:
= ∫ (8 sec^2θ dθ) / (64 + 64 tan^2θ)^2
= ∫ (8 sec^2θ dθ) / (64 (1 + tan^2θ))^2
Since sec^2θ = 1 + tan^2θ, we can simplify further:
= ∫ (8 sec^2θ dθ) / (64 sec^4θ)^2
= ∫ 8 sec^2θ dθ / 4096 sec^8θ
Step 3: Simplify the integral:
Now, we can simplify the integral by canceling out common terms:
= (1/512) ∫ sec^2θ / sec^8θ dθ
= (1/512) ∫ sec^2θ / sec^2θ^4 dθ
= (1/512) ∫ dθ / (sec^2θ)^3
Since sec^2θ = 1/cos^2θ, we can simplify further:
= (1/512) ∫ cos^6θ dθ
Step 4: Evaluate the integral:
The integral of cos^6θ can be computed using the reduction formula, which states that:
∫ cos^nθ dθ = (1/n) cos^(n-1)θ sinθ + [(n-1)/n] ∫ cos^(n-2)θ dθ
Using this formula, we can simplify the integral:
∫ cos^6θ dθ = (1/6) cos^5θ sinθ + (5/6) ∫ cos^4θ dθ
= (1/6) cos^5θ sinθ + (5/6) [(1/4) cos^3θ sinθ + (3/4) ∫ cos^2θ dθ]
= (1/6) cos^5θ sinθ + (5/6) [(1/4) cos^3θ sinθ + (3/4) [(1/2) cosθ sinθ + (1/2) ∫ dθ]]
= (1/6) cos^5θ sinθ + (5/6) [(1/4) cos^3θ sinθ + (3/8) cosθ sinθ + (3/8) θ] + C
This expression gives the value of the indefinite integral. Remember to add the constant of integration (C) at the end.
Therefore, the indefinite integral of dx / (64 + x^2)^2 is:
(1/6) cos^5θ sinθ + (5/6) [(1/4) cos^3θ sinθ + (3/8) cosθ sinθ + (3/8) θ] + C
However, we need to convert the result back to the variable x, so let's substitute back:
(1/6) cos^5θ sinθ + (5/6) [(1/4) cos^3θ sinθ + (3/8) cosθ sinθ + (3/8) θ] + C
= (1/6) (1 - x^2/64)^5 (x/8) + (5/6) [(1/4) (1 - x^2/64)^3 (x/8) + (3/8) (1 - x^2/64) (x/8) + (3/8) θ] + C
This is the final result of the indefinite integral of dx / (64 + x^2)^2 in terms of x.