Find the intergral of (1+x)/sqrt(3-x^2)
To find the integral of the function (1+x)/sqrt(3-x^2), we can use a technique called trigonometric substitution. This method involves substituting x with a trigonometric expression that simplifies the integral.
Let's start by making the substitution x = sqrt(3)sin(θ). Now, let's find the derivative of x with respect to θ:
dx/dθ = sqrt(3)cos(θ)
Next, let's express dx in terms of dθ:
dx = sqrt(3)cos(θ)dθ
Now, substitute the values of x and dx back into the integral:
∫(1+x)/sqrt(3-x^2) dx = ∫(1 + sqrt(3)sin(θ))/(sqrt(3)cos(θ)) * sqrt(3)cos(θ) dθ
= sqrt(3)∫(1 + sqrt(3)sin(θ))/(cos(θ)) dθ
Simplifying the numerator:
= sqrt(3)∫(sqrt(3)sin(θ) + 1)/(cos(θ)) dθ
= sqrt(3)∫tan(θ) + sec(θ) dθ
The integral now simplifies to:
= sqrt(3)∫tan(θ) dθ + sqrt(3)∫sec(θ) dθ
To solve these new integrals, we need to express tan(θ) and sec(θ) in terms of sin(θ) and cos(θ). Using the trigonometric identities:
tan(θ) = sin(θ)/cos(θ)
sec(θ) = 1/cos(θ)
Substituting these values back into the integral:
= sqrt(3)∫(sin(θ)/cos(θ)) dθ + sqrt(3)∫(1/cos(θ)) dθ
The first integral can be solved by a simple substitution, by letting u = cos(θ):
= sqrt(3)∫u^-1 du
Integrating u^-1 gives us:
= sqrt(3)ln|u| + C1
Next, let's solve the second integral:
= sqrt(3)∫(1/cos(θ)) dθ
This integral can be rewritten as:
= sqrt(3)∫sec(θ) dθ
Using the identity sec(θ) = 1/cos(θ), we get:
= sqrt(3)ln|sec(θ) + tan(θ)| + C2
So, the final solution is:
= sqrt(3)ln|cos(θ)| + sqrt(3)ln|sec(θ) + tan(θ)| + C
To express the solution in terms of x, we need to substitute back x = sqrt(3)sin(θ):
= sqrt(3)ln|cos(θ)| + sqrt(3)ln|sec(θ) + tan(θ)| + C
Now, using the identity cos(θ) = sqrt(1 - sin^2(θ)), we can insert x back into the solution:
= sqrt(3)ln|cos(arcsin(x/sqrt(3)))| + sqrt(3)ln|sec(arcsin(x/sqrt(3))) + tan(arcsin(x/sqrt(3)))| + C
Finally, simplifying the expression:
= sqrt(3)ln|cos(arcsin(x/sqrt(3)))| + sqrt(3)ln|1/cos(arcsin(x/sqrt(3))) + x/sqrt(3)/cos(arcsin(x/sqrt(3)))| + C
And that is the integral of (1+x)/sqrt(3-x^2).