
Integral
I am finding it very hard to solve this integration problem. Can some one explain the process step by step so that I will be able to do futher problems by myself.
Heres the question:
$\displaystyle \int\frac{1}{x\sqrt{49In^2(x)}}dx$
I know the solution but don't know how to get to it!
I hope this question is useful to others too.
Thanks

I assume you mean lnē(x) instead of "In". You should immediately see that ln(x)'s derivative, 1/x, is present too which makes the following substitution useful:
$\displaystyle
y = \ln x \Leftrightarrow dy = \frac{1}{x}dx
$
This converts the integral to:
$\displaystyle
\int {\frac{1}{{\sqrt {4  9y^2 } }}dy} = \int {\frac{1}{{\sqrt {4\left( {1  \frac{9}{4}y^2 } \right)} }}dy} = \int {\frac{1}{{2\sqrt {1  \frac{9}{4}y^2 } }}dy}
$
Now I created "sqrt(1...)" in the denominator. If we can get that "..." to be a square, then the inverse sine should ring a bell! So let's stick that 9/4 within the square and adjust the dx (instead of another substitution):
$\displaystyle
\frac{1}{2}\int {\frac{1}{{\sqrt {1  \left( {\frac{{3y}}{2}} \right)^2 } }}dy} = \frac{1}{2}\frac{2}{3}\int {\frac{1}{{\sqrt {1  \left( {\frac{{3y}}{2}} \right)^2 } }}d\left( {\frac{{3y}}{2}} \right)}
$
Now this is a standard integral, which yeilds arcsin of the integration variable.
$\displaystyle
\frac{1}{3}\arcsin \left( {\frac{{3y}}{2}} \right) + C \to \frac{1}{3}\arcsin \left( {\frac{3}{2}\ln x} \right) + C
$
I hope this is clear :)

re:
Hi TD! :)
You make maths seem so simple lol! Genius you got the right answer.
Well I get what you have done until the 3rd line where do you get the 2/3 from ? (the 2/3 multiplying with the 1/2)
Thanks

Actually, that is a "hidden substitution". I change the integration variable from y to 3y/2 which is allowed, if you correct with a factor 2/3 since d(y) = 2/3*3/2 d(y) = 2/3 d(3y/2). If you're uncomfortable with that, just use another substitution: 3y/2 = u <=> 3/2 dy = du <=> dy = 2/3 du, and there appears the 2/3.

Oh right! Thanks
I think I will better stick with another substitution! As it is probably easier for you as you can probably do it simply by looking at it.
Thanks for that.
Kind Regards
dadon

No problem, let us know if it doesn't work out for you :)