How do we integrate {[1/(√(1-x^2) )]*[ln|(2x+3)(√(1-x^2))] } ?
Is there any suitable substitution or do we have to do this by parts?
I don't have any idea to proceed on.
Can we use the substitution t= ln|(2x+3)*(1-x^2)^(1/2)|?
Will that make this simpler?
how about this:
x = sinθ
√(1-x^2) = cosθ
dx = cosθ dθ
so, dθ = dx/cosθ = dx/√(1-x^2)
Then the integrand is
ln((2sinθ+3)cosθ) dθ
Unfortunately, that doesn't integrate any better. We can't use the same trick from before, since we don't have a definite integral. If we did, it might loosen things up a bit.
What if the limits are 0-1?
To integrate the given expression, we can use the technique of integration by parts. Let's start by applying the integration by parts formula:
∫u * dv = uv - ∫v * du
Let's assign the following:
u = ln|(2x+3)(√(1-x^2))|
dv = 1/(√(1-x^2))
Now, we need to find du and v:
To find du, we can take the derivative of u with respect to x:
du/dx = 1/[(2x+3)(√(1-x^2))] * d/dx [(2x+3)(√(1-x^2))]
To find v, we need to integrate dv. In this case, the integral of dv represents arcsin(x), which is the inverse sine function.
Now, let's find du and v step by step.
1. Find du:
To get du/dx, we can use the product rule of differentiation:
du/dx = 1/[(2x+3)(√(1-x^2))] * [d/dx(2x+3)(√(1-x^2))]
By applying the product rule, we have:
du/dx = 1/[(2x+3)(√(1-x^2))] * [2(√(1-x^2)) + (2x+3)(-2x/(√(1-x^2)))]
Simplifying further:
du/dx = 2/√(1-x^2) - 2x(2x+3)/[(2x+3)(√(1-x^2))]
Simplifying and canceling out common terms:
du/dx = 2/√(1-x^2) - 4x^2/(√(1-x^2))
2. Find v:
Integrating dv, which is 1/√(1-x^2), gives us arcsin(x).
Now, we have found du and v. We can substitute these values into the integration by parts formula:
∫[1/(√(1-x^2))] * [ln|(2x+3)(√(1-x^2))|] dx
= uv - ∫v * du
= ln|(2x+3)(√(1-x^2))| * arcsin(x) - ∫arcsin(x) * [2/√(1-x^2)-4x^2/(√(1-x^2))] dx
At this point, we have an integral of arcsin(x) times some terms. To simplify further, we can split the integral into two separate integrals:
∫arcsin(x) * [2/√(1-x^2)-4x^2/(√(1-x^2))] dx
= ∫arcsin(x) * (2/√(1-x^2)) dx - ∫arcsin(x) * (4x^2/(√(1-x^2))) dx
The first integral, with arcsin(x) * (2/√(1-x^2)), could be evaluated by using integration by parts once again with u = arcsin(x) and dv = (2/√(1-x^2))dx.
For the second integral, we can simplify it further by using the trigonometric identity: sin^2(x) + cos^2(x) = 1, which rearranges to cos^2(x) = 1 - sin^2(x). Substituting this identity, the integral becomes more manageable.
The above steps outline the approach to integrate the given expression using integration by parts and additional simplifications with trigonometric identities. The detailed calculations can be performed to obtain the final result.