# Thread:

1. ## Indefinite integral

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

2. 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

3. 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.

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

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

5. I think the answer is:

2sqrt(arcsinx) + c

6. 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?

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

$2\sqrt{\arcsin{x}} + C$
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9288584589958191, "perplexity_flag": "middle"}