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..

<br /> x = cos(u)~;~du = -sin(u)<br />
also,
<br /> x^2 = cos^2(u)<br />

Therefore we have,

<br /> \int\frac{cos^{-1}(x)}{x^2}dx = I = \int\frac{u sin(u)}{cos^{2}(u)}du<br />

<br /> I = -\int u tan(u)sec(u)du<br />

<br /> I = -\left[u sec(u) - \int \frac{du}{du} \int sec(u)tan(u)du du\right]<br />

<br /> I = -\left[u sec(u) - \int sec(u)du\right]<br />

<br /> I = log_e|sec(u) + tan(u)| - u sec(u) + C<br />

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