Indefinite integral

Mar 2010
31
0
Find the following indefinite integral, identifying any rules of calculus that you use:

\(\displaystyle \int1/\sqrt{(1-x^2)*\arcsin(x)}\)

I think I need to use integration by substitution, so have rearranged to give:

\(\displaystyle \int(1-x^2)^{\-1/2} * (sin(x))^{1/2}\)

I am not quite sure what to do next. The formula I have for integration by substitution is:

\(\displaystyle \int{f(g(x))g'(x)dx} = \int{f(u)du}\), where u = g(x)

Would I then get

f(x) = \(\displaystyle (1-x^2)^{\-1/2}\)

and

g(x) = \(\displaystyle (sin(x))^{1/2}\), so u = -cosx

or do I need to integrate both parts first as they are both composite integrals and then use the above formula on the answers I get? (Headbang)

Thanks for any help in advence
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Find the following indefinite integral, identifying any rules of calculus that you use:

\(\displaystyle \int1/\sqrt{(1-x^2)*\arcsin(x)}\)

\(\displaystyle \displaystyle \int \frac{1}{\sqrt{\arcsin{x}}} \cdot \frac{1}{\sqrt{1-x^2}} \, dx\)

\(\displaystyle u = \arcsin{x}
\)

\(\displaystyle \displaystyle du = \frac{1}{\sqrt{1-x^2}} \, dx\)

substitute ...

\(\displaystyle \displaystyle \int \frac{1}{\sqrt{u}} \, du
\)

finish it
 
  • Like
Reactions: cozza
Jul 2010
58
27
Brazil
I'll give you a huge hint. You will have to use the u-substitution, this identity:

\(\displaystyle \displaystyle{\frac{d}{dx} arcsin(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2}}\)

and the integral rewritten as: \(\displaystyle \displaystyle{\int\frac{1}{\sqrt{(1-x^2)\cdot arcsin(x)}}dx=\int (1-x^2)^{-1/2}\cdot(arcsin(x))^{-1/2}dx}\)

This should work.
 
  • Like
Reactions: cozza
Mar 2010
31
0
Hi, thanks for the help. Can I just check that the answer I have is right:

\(\displaystyle 1/(\sqrt{1-x^2})+c\)
 
Mar 2010
31
0
I think the answer is:

2sqrt(arcsinx) + c
Hi, thanks for the help. Can I just check that the answer I have is right:

\(\displaystyle 1/(\sqrt{1-x^2})+c\)
On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

\(\displaystyle 1/(1-x^2)\) ?

Where does the 2 come from? Have I missed something?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

\(\displaystyle 1/(1-x^2)\) ?

Where does the 2 come from? Have I missed something?
yes, you have.


as stated earlier ...

\(\displaystyle \displaystyle \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C
\)

back substitute \(\displaystyle u = \arcsin{x}\) ...

\(\displaystyle 2\sqrt{\arcsin{x}} + C\)