Another indefinite integral question?

[uman]

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

 

[Rainbow Child]

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

 

[rohanprabhu]

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

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.

 

[arildno]

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

 

[rohanprabhu]

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

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

 

[arildno]

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

 

[rohanprabhu]

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