Question

Question #204ba

Answer 1

`1/3*arc sin(3/2*lnx)+C.`

Explanation:

Let, `I=intdx/(xsqrt(4-9(lnx)^2).`

Substitute `lnx=t" so that, "dx/x=dt.`

`:. I=intdt/sqrt(4-9t^2).`

`=1/3*arc sin((3t)/2).`

`rArr I=1/3*arc sin(3/2*lnx)+C.`

Answer 2

`int (dx)/(xsqrt[4-9(Lnx)^2])=1/3*arcsin((3Lnx)/2)+C`

Explanation:

`int (dx)/(xsqrt[4-9(Lnx)^2])`

=`1/3*int (3dx)/(xsqrt[4-9(Lnx)^2])`

After using `3Lnx=2sinu` and `(3dx)/x=2cosu*du` transforms, this integral became

`1/3*int (2cosu*du)/sqrt[4-4(sinu)^2]`

=`1/3*int (2cosu*du)/sqrt[4(cosu)^2]`

=`1/3*int (2cosu*du)/(2cosu)`

=`1/3*int du`

=`1/3*u+C`

After using `3Lnx=2sinu` and `u=arcsin((3Lnx)/2)` inverse transforms, I found

`int (dx)/(xsqrt[4-9(Lnx)^2])=1/3*arcsin((3Lnx)/2)+C`

Answer 3

This is done by the substitution method.

Explanation:

Let `lnx` be t.
Now, differentiate this equation on both sides.
`1/x dx` = `dt`
Now you can apply the formula of `int 1/sqrt(a^2-x^2)` = `sin^-1(x/a)`
and later put t = `ln x` back in the equation.