Use the substitution x=3tan() to evaluate the indefinite integral 93dx/(x^2sprt(x^2+9))
correction: use the substitution x=3tan(theta)
so, try it. where do you get stuck?
I'm not sure if this is right but after substituting and simplifying I get to:
integral (93/9tan^2theta*3sectheta)*3sec^2theta dtheta
It took me your trig substitution followed by a u substitution to get 93squrt(x^2+9)/9x+C
To evaluate the indefinite integral ∫ 93dx / (x^2 √(x^2 + 9)), we can use the substitution x = 3tan(θ).
Let's start by finding dx in terms of dθ:
Differentiating both sides of the substitution equation x = 3tan(θ) with respect to θ, we get:
dx = 3sec^2(θ) dθ
Now we substitute these expressions into the integral:
∫ 93dx / (x^2 √(x^2 + 9)) = ∫ 93(3sec^2(θ) dθ) / ((3tan(θ))^2 √((3tan(θ))^2 + 9))
Simplifying this expression gives us:
∫ 93(3sec^2(θ) dθ) / (9tan^2(θ) √(9tan^2(θ) + 9))
= ∫ (279sec^2(θ) dθ) / (9tan^2(θ) √(9tan^2(θ) + 9))
Simplifying further, we have:
∫ (31sec^2(θ) dθ) / (tan^2(θ) √(tan^2(θ) + 1))
To proceed, we can rewrite sec^2(θ) as 1 + tan^2(θ), and tan^2(θ) as (sec^2(θ) - 1):
∫ (31(1 + tan^2(θ)) dθ) / (tan^2(θ) √(tan^2(θ) + 1))
= ∫ (31(1 + (sec^2(θ) - 1)) dθ) / ((sec^2(θ) - 1) √((sec^2(θ) - 1) + 1))
= ∫ (31sec^2(θ) dθ) / (sec^2(θ) √sec^2(θ))
= 31 ∫ (sec^2(θ) dθ) / sec^2(θ)
= 31 ∫ dθ
Finally, integrating dθ gives us:
= 31θ + C
Therefore, the indefinite integral ∫ 93dx / (x^2 √(x^2 + 9)) evaluates to 31θ + C, where θ is the inverse tangent (arctan) of (x/3).