Another indefinite integral question?

1. Jan 13, 2008

uman

How to evaluate $$\int{\frac{\arccos{x}}{x^2}\,dx}$$?

2. Jan 13, 2008

Rainbow Child

Try the substition
$$x=\cos u,\,d\,x=-\sin u\,d\,u$$
and then integrate by parts.

3. Jan 15, 2008

rohanprabhu

Perfect..

$$x = cos(u)~;~du = -sin(u)$$
also,
$$x^2 = cos^2(u)$$

Therefore we have,

$$\int\frac{cos^{-1}(x)}{x^2}dx = I = \int\frac{u sin(u)}{cos^{2}(u)}du$$

$$I = -\int u tan(u)sec(u)du$$

$$I = -\left[u sec(u) - \int \frac{du}{du} \int sec(u)tan(u)du du\right]$$

$$I = -\left[u sec(u) - \int sec(u)du\right]$$

$$I = log_e|sec(u) + tan(u)| - u sec(u) + C$$

Then, you can replace the value of u with $cos(x)$ to get the required integral.

4. Jan 15, 2008

arildno

Blunder there, you are to substitute $u=\cos^{-1}(x)$

Last edited: Jan 15, 2008
5. Jan 15, 2008

rohanprabhu

$u = \cos^{-1}(x)~;~x = \cos(u)$.. it's the same thing within a restricted domain.

6. Jan 15, 2008

arildno

You wrote substitute u with cos(x), which is wrong.

7. Jan 15, 2008

rohanprabhu

ohh.. u mean in the last part.. yeah.. i get it now... ur right :D