How do you integrate #int x^2/(sqrt(x^2-4))dx# using trigonometric substitution?

Let, #I=intx^2/sqrt(x^2-4)dx#.

We subst. #x=2secy," so that, "dx=2secytanydy#.

#:. I=int(4sec^2y)/sqrt(4sec^2y-4)*2secytanydy#,

#=int(4sec^2y)/(2tany)*2secytanydy#,

#=4intsec^3ydy#,

#=4intsecy*sec^2ydy#,

#=4intsqrt(tan^2y+1)sec^2ydy#.

Next, we use the substn. #tany=t :. sec^2ydy=dt#.

#:. I=4intsqrt(t^2+1)dt#,

#=4{t/2sqrt(t^2+1)+1/2ln|t+sqrt(t^2+1)|}#,

#=2{tanysecy+ln|tany+secy|}......[because, t=tany]#.

Returning to #secy=x/2," so that, "tany=sqrt(x^2/4-1)#, we have,

#I=2{x/2*sqrt(x^2-4)/2+ln|x/2+sqrt(x^2-4)/2|}#,

#rArr I=x/2sqrt(x^2-4)+2ln|x/2+sqrt(x^2-4)/2|+C, or, #

# I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c, c=C-2ln2#.

However, the integral can easily be dealt with without

using the substn. as shown below :

#I=intx^2/sqrt(x^2-4)dx=int{(x^2-4)+4}/sqrt(x^2-4)dx#,

#int{(x^2-4)/sqrt(x^2-4)+4/sqrt(x^2-4)}dx#,

#=intsqrt(x^2-4)dx+4int1/sqrt(x^2-4)dx#,

#={x/2sqrt(x^2-4)-4/2ln|x+sqrt(x^2-4)|}+4ln|x+sqrt(x^2-4)|#.

#I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+C_1#, as before!

Enjoy Maths.!

We have,
#color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-(a^2)/2ln|x+sqrt(x^2-a^2)|#
#color(red)((2)int1/sqrt(x^2-a^2)dx=ln|x+sqrt(x^2-4)|+c#
Hence,
#I=int(x^2)/sqrt(x^2-4)dx=int((x^2-4)+4)/sqrt(x^2-4)dx#
#I=int(x^2-4)/sqrt(x^2-4)dx+int4/sqrt(x^2-4)dx#
#=intsqrt(x^2-4) dx+4int1/sqrt(x^2-4)dx#
#=intsqrt(x^2-2^2)dx+4int1/sqrt(x^2-2^2)dx#
Using (1) and (2)
#I=x/2sqrt(x^2-2^2)-(2^2)/2ln|x+sqrt(x^2-2^2)|+4ln|x+sqrt(x^2-2^2)|#
#=x/2sqrt(x^2-4)-2ln|x+sqrt(x^2-4)|+4ln|x+sqrt(x^2-4)|+c#
#I=x/2sqrt(x^2-4)+2ln|x+sqrt(x^2-4)|+c#

For Trigonometric substitution proceed according to Hint given
above