I'm not sure what to substitute in this question.
Evaluate the integral
Square root (tan^-1(x)) / (x^2+1)
To evaluate the integral of √(tan^(-1)(x))/(x^2+1), we can use a technique called integration by substitution.
The first step is to choose an appropriate substitution. In this case, let's substitute u = tan^(-1)(x).
Next, we need to find the derivative of u with respect to x. Taking the derivative of both sides of the equation u = tan^(-1)(x), we get du/dx = 1/(1+x^2). Rearranging this equation, we have dx = du/(1+x^2).
Now, we substitute these values into the integral:
∫(√(tan^(-1)(x))/(x^2+1)) dx
= ∫ (√u)/(1+x^2) du
= ∫ (√u)/(1+tan^2(u)) du
= ∫ (√u)/(sec^2(u)) du
= ∫ sin(u)√u du
We can simplify the expression further by using the identity sin^2(u) = (1-cos(2u))/2.
∫ sin(u)√u du
= ∫ √u sin(u) du
= ∫ √u (2sin^2(u/2)) du
= 2 ∫ (sin(u/2))^2 √u du
Now, we have a more familiar form involving trigonometric functions.
At this point, we can apply another substitution, such as v = u/2.
Taking the derivative of v with respect to u, we get dv/du=1/2. Rearranging this equation, we have du = 2dv.
Substituting these values into the integral:
2 ∫ (sin(u/2))^2 √u du
= 2 ∫ (sin(v))^2 √(2v) 2dv
= 4 ∫ sin^2(v) √(2v) dv
= 4 ∫ (1-cos(2v))/2 √(2v) dv
= 2 ∫ (1-cos(2v)) √(2v) dv
Expanding the integral:
= 2 ∫ √(2v) dv - 2 ∫ cos(2v) √(2v) dv
The first integral can be easily evaluated as:
= 2 * (2/3) (2v)^(3/2) + C
= (4/3) (2v)^(3/2) + C
= (8/3) v^(3/2) + C
Now, we need to evaluate the second integral:
- 2 ∫ cos(2v) √(2v) dv
This integral can be computed using integration by parts or by using a different substitution. The particular method depends on your preference and familiarity.
After integrating both terms, you can substitute back the original variable u and x to get the final result.