Solve the integral: (1-x^2)^(3/2)/x^2
This is what I have so far. I used trig substitution.
x=sin theta
dx=cos theta dtheta
[1-(sin^2 theta)^(3/2)]*[cos theta dtheta]/(sin^2 theta)
[(cos^3 theta)]*[cos theta dtheta]/(sin^2 theta)
(cos^4 theta)/(sin^2 theta)
so far so good
cos^4θ/sin^2θ = (1-2sin^2θ+sin^4θ)/sin^2θ
= csc^2θ - 2 + sin^2θ
= csc^2θ - 2 + (1-cos2θ)/2
all of those are simple to integrate.
Steve thx for answering at that time
To solve the integral of (1-x^2)^(3/2)/x^2, you made a good start by using trigonometric substitution. You substituted x = sin(theta) and dx = cos(theta) dtheta.
Now, you have the integral:
[(cos^4 theta)/(sin^2 theta)] dtheta
To simplify this expression, you can use the trigonometric identity sin^2 theta + cos^2 theta = 1. Rearranging it, sin^2 theta = 1 - cos^2 theta.
Substituting this into the integral, you get:
[(cos^4 theta)/((1-cos^2 theta))] dtheta
Next, you can simplify further by factoring out a common factor:
[(cos^4 theta)/(sin^2 theta * (1-cos^2 theta))] dtheta
Since sin^2 theta = 1 - cos^2 theta, you can rewrite the denominator as:
[(cos^4 theta)/((1 - cos^2 theta) * cos^2 theta)] dtheta
Simplifying the denominator gives:
[(cos^4 theta)/(cos^2 theta * sin^2 theta)] dtheta
Now, you can simplify by canceling out the common factor cos^2 theta:
[(cos^2 theta)/(sin^2 theta)] dtheta
Using a trigonometric identity, tan^2 theta = sin^2 theta / cos^2 theta, you can rewrite the integral as:
[(cos^2 theta)/(1 - cos^2 theta)] dtheta
Finally, you can use a substitution u = cos^2 theta, which means du = -2sin theta cos theta dtheta.
Rearranging the terms, you have -du = 2sin theta cos theta dtheta, so you can rewrite the integral as:
-[(1/(1 - u))] du
To solve this integral, you can integrate -(1/(1 - u)) with respect to u:
- ln|1 - u| + C
Finally, substitute u back in terms of theta to get the final answer:
- ln|1 - cos^2 theta| + C
Note: Make sure to check the appropriate limits of integration and adjust the sign of the logarithm accordingly for the final answer.