1. ## Trig-sub problem

$\int{\frac{(1-x^2)^{3/2}}{x^2}}$

After my Trig sub , i get $\frac{1}{x}+2arccos(x)-\frac{x^3}{3}+C$

when i differentiate that, i do not get my initial integral which is a problem , can someone do this integral step by step please

2. Originally Posted by Larrioto
$\int{\frac{(1-x^2)^{3/2}}{x^2}}$

After my Trig sub , i get $\frac{1}{x}+2arccos(x)-\frac{x^3}{3}+C$

when i differentiate that, i do not get my initial integral which is a problem , can someone do this integral step by step please

Here is how I approached it
x =cos(t)

dx = - sin(t)dt

$-\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}$

$-\int{\frac{(sin^{4}(t)) }{cos^2(t)}}$

$-\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}$

$-\int{tan^{2}(t)-sin^2(t) }$

$- tan(t) + t + \int{\frac{1-cos(2t)}{2}}$

I wont go further before you try it

3. $-\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}$

this is false, because the exponant is not 3, but 3/2

4. Originally Posted by Larrioto
$-\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}$

this is false, because the exponant is not 3, but 3/2
$(1-\cos^2(x))^{3/2}=(\sin^2(x))^{3/2}=\sin(x)^{2 \times 3/2}=\dots$

5. The error is here.....

Here is how I approached it
x =cos(t)

dx = - sin(t)dt

$-\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}$

$-\int{\frac{(sin^{4}(t)) }{cos^2(t)}}$

$-\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}$

$-\int{tan^{2}(t) TIMES sin^2(t) }$

$- tan(t) + t + \int{\frac{1-cos(2t)}{2}}$

I wont go further before you try it
It's times and not subtraction $-\int{\frac{sin^{4}(t)}{cos^2(t)}} =-\int{\frac{sin^{2}(t) }{cos^2(t)}}\sin^2t$

maybe try the two half angle formulas here instead of going with tangent times sine

Here is how I approached it
x =cos(t)

dx = - sin(t)dt

$-\int{\frac{(sin(t))^{3}\times sin(t) }{cos^2(t)}}$

$-\int{\frac{(sin^{4}(t)) }{cos^2(t)}}$

$-\int{\frac{(sin^{2}(t))(1-cos^2(t)) }{cos^2(t)}}$

$-\int{\frac{(sin^{2}(t)\times(1) }{cos^2(t)}}-\frac{sin^2(t)\times cos^2(t)}{cos^2(t)}$

$-\int{tan^{2}(t)-sin^2(t) }$

$- tan(t) + t + \int{\frac{1-cos(2t)}{2}}$

I wont go further before you try it
*I dont understand it

7. It looks fine. I guess through trig the integrand can be either.
I figured you had to have a mistake, but they both seem ok.
Now you need to get back to the x.

$- tan(t) + t + \int{\frac{1-cos(2t)}{2}}$
$-tan(t) + \frac{3t}{2}-sin(2t)/4$

$cos(t)=x$

$tan(t) = \sqrt{\frac{1}{cos^2(t)} - 1}=\sqrt{\frac{1}{x^2} - 1}$

$t =cos^{-1}(x)$

$sin(2t)/4 = \frac{ sin(t)cos(t)}{2} = \frac{\sqrt{(1-cos^2(t))}\times cos(t)}{2}$ $= \frac{\sqrt{(1-x^2)}\times x}{2}$