# Indefinite integral

#### cozza

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
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

• cozza

#### bondesan

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.

• cozza

#### cozza

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

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

#### Also sprach Zarathustra

2sqrt(arcsinx) + c

#### cozza

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
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 $$\displaystyle u = \arcsin{x}$$ ...

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