Question #204ba

3 Answers
Ratnaker Mehta
Nov 25, 2017

# 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.#

Cem Sentin
Nov 25, 2017

#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#

T1LT
Nov 25, 2017

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.

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