Indefinite integral

**cozza** · July 26th 2010, 09:05 AM

Find the following indefinite integral, identifying any rules of calculus that you use:

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

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

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

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

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

Would I then get

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

and

g(x) = $(sin(x))^{1/2}$ , 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?

Thanks for any help in advence

**skeeter** · July 26th 2010, 09:21 AM

Originally Posted by cozza

Find the following indefinite integral, identifying any rules of calculus that you use:

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

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

$u = \arcsin{x}<br />$

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

substitute ...

$\displaystyle \int \frac{1}{\sqrt{u}} \, du<br />$

finish it

**bondesan** · July 26th 2010, 09:33 AM

I'll give you a huge hint. You will have to use the u-substitution, this identity:

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

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

This should work.

**cozza** · July 27th 2010, 09:11 AM

Hi, thanks for the help. Can I just check that the answer I have is right:

$1/(\sqrt{1-x^2})+c$

**Also sprach Zarathustra** · July 27th 2010, 09:19 AM

I think the answer is:

2sqrt(arcsinx) + c

**cozza** · July 27th 2010, 09:26 AM

Originally Posted by Also sprach Zarathustra

I think the answer is:

2sqrt(arcsinx) + c

Originally Posted by cozza

Hi, thanks for the help. Can I just check that the answer I have is right:

$1/(\sqrt{1-x^2})+c$

On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

$1/(1-x^2)$ ?

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

**skeeter** · July 28th 2010, 09:30 AM

Originally Posted by cozza

On rechecking my answer I don't think the square root should be in there. I think (hope) it is:

$1/(1-x^2)$ ?

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

yes, you have.

as stated earlier ...

$\displaystyle \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C<br />$

back substitute $u = \arcsin{x}$ ...

$2\sqrt{\arcsin{x}} + C$

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