Integrate sqrt(16+x^2) dx
let x = 4tanθ
√(16+x^2) = 4secθ
dx = 4sec^2θ dθ
it gets much simpler.
To find the integral of √(16+x^2) dx, we can use a trigonometric substitution. Let's substitute x = 4sinθ.
First, let's find dx in terms of dθ:
dx = 4cosθ dθ
Next, let's substitute the values of x and dx in terms of θ into the integral:
∫√(16+x^2) dx = ∫√(16+(4sinθ)^2) * 4cosθ dθ
= ∫√(16+16sin^2θ) * 4cosθ dθ
= ∫√(16(1+sin^2θ)) * 4cosθ dθ
= ∫√(16cos^2θ + 16sin^2θ) * 4cosθ dθ
= ∫√(16(cos^2θ + sin^2θ)) * 4cosθ dθ
= ∫√(16) * 4cosθ dθ
= ∫4cosθ dθ
Since ∫cosθ dθ = sinθ, the integral becomes:
= 4∫cosθ dθ
= 4sinθ + C
Finally, we need to substitute the value of θ back into the equation. Recall that we originally set x = 4sinθ, so we need to find θ in terms of x.
x = 4sinθ
sinθ = x/4
θ = arcsin(x/4)
Substituting the value of θ back into the solution:
= 4sin(arcsin(x/4)) + C
= 4(x/4) + C
= x + C
Therefore, the indefinite integral of √(16+x^2) dx is x + C, where C is the constant of integration.