How do you integrate #int -x^2/sqrt(144+x^2)dx# using trigonometric substitution?

3 Answers
Jul 16, 2018

#=>I=72ln|x+sqrt(x^2+144)|-x/2sqrt(x^2+144)+C#

Explanation:

Here ,

#I=-intx^2/sqrt(x^2+144)dx#

Subst. #color(blue)(x=12tanu=>dx=12sec^2udu and #

#color(blue)(tanu=x/12=>secu=sqrt(1+tan^2u)=sqrt(1+x^2/144)#

#=>color(blue)(secu=sqrt(144+x^2)/12#

So,

#I=-int(144tan^2u)/sqrt(144tan^2u+144)*12sec^2udu#

#=>I=-144int(tan^2u)/(12secu)12sec^2udu#

#=>I=-144intsecutan^2udu#

#=>I=-144*I_1----to(A)#

#Where , I_1=intsecutan^2udu...to(B)#

#=>I_1=intsecu(sec^2u-1)du...to[becausesec^2theta- tan^2theta=1]#

#=>I_1=intsecusec^2udu-intsecudu#

Using Integration by Parts in first integral

#I_1#=#{secuintsec^2u-int(secutanuintsec^2udu)du}-intsecudu#

#I_1#=#{secu*tanu-intsecutanu*tanudu}-ln|secu+tanu|#

#=>I_1=secutanu-intsecutan^2udu-ln|secu+tanu|#

#=>I_1=secutanu-I_1-ln|secu+tanu|+c...to[use ,(B)]#

#=>2I_1=secutanu-ln|secu+tanu|+c#
Subst. back , #color(blue)(tanu=x/12 and secu=sqrt(144+x^2)/12#

#2I_1=(sqrt(144+x^2))/12*x/12-ln|sqrt(144+x^2)/12+x/12|+c#

#=>2I_1=x/144sqrt(144+x^2)-{ln|sqrt(144+x^2)+x|-ln12}+c#

#=>I_1=1/2*x/144sqrt(144+x^2)-1/2ln|sqrt(144+x^2)+x|+C'#

#where , C'=1/2(c-ln12)#

Now ,from #(A)# we get

#I#=#-144{1/2*x/144sqrt(144+x^2)-1/2ln|x+sqrt(144+x^2)|}+C#

#I=-1/2xsqrt(144+x^2)+72ln|x+sqrt(x^2+144)|+C#

#=>I=72ln|x+sqrt(x^2+144)|-x/2sqrt(x^2+144)+C#

Jul 16, 2018

The answer is #=-6xsqrt(1+x^2/144)+72ln(|x/12+sqrt(1+x^2/144)|)+C#

Explanation:

The integral is

#I=int-(x^2dx)/sqrt(144+x^2)#

Rewrite it as

#I=int-((x^2+144-144)dx)/sqrt(144+x^2)#

#=int(144dx)/(sqrt(144+x^2))-int((144+x^2)dx)/sqrt(144+x^2)#

#=int(144dx)/(sqrt(144+x^2))-intsqrt(144+x^2)dx#

#=I_1+I_2#

Compute the #2nd# integral by substitution

Let #x=12tantheta#

#dx=12sec^2theta d theta#

#sqrt(144+x^2)=sqrt(144+144tan^2theta)=12sectheta#

Therefore,

#I_2=intsqrt(144+x^2)dx=144intsec^3thetad theta=144intsec^2thetasecthetad theta#

Perform this integration by parts

#u=sectheta#, #=>#, #u'=secthetatantheta#

#v'=sec^2theta#, #=>#, #v=tantheta#

Therefore,

#144intsec^3thetad theta=144secthetatantheta-144intsecthetatan^2theta d theta#

#=144secthetatantheta-144int(sec^2theta-1)secthetad theta#

#=144secthetatantheta-144intsec^3theta d theta+144 intsecthetad theta#

#288intsec^3theta d theta=144secthetatantheta+144intsecthetad theta#

#=144secthetatantheta+144ln(|sectheta+tantheta|)#

#144intsec^3theta d theta=72secthetatantheta+72ln(|sectheta+tantheta|)#

#I_2=72*x/12sqrt(1+x^2/144)+72ln(|x/12+sqrt(1+x^2/144)|)#

Compute the first integral

#I_1=int(144dx)/(sqrt(144+x^2))=144int(dx)/sqrt(144+x^2)#

Let #u=x/12#, #=>#, #du=1/12dx#

#I_1=144int(12du)/(12sqrt(1+u^2)#

#=144int(du)/sqrt(1+u^2)#

Let #u=tanv#, #=>#, #du=sec^2vdv#

#sqrt(1+u^2)=sqrt(1+tan^2v)=secv#

#I_1=144int(sec^2vdv)/(secv)=144intsecvdv#

#=144ln(|secv+tanv|)#

#=144ln(sqrt(1+u^2)+u)#

#=144ln(sqrt(1+x^2/144)+x/12)#

Putting it alltogether

#I=-6xsqrt(1+x^2/144)-72ln(x/12+sqrt(1+x^2/144))+144ln(sqrt(1+x^2/144)+x/12)#

#=-6xsqrt(1+x^2/144)+72ln(|x/12+sqrt(1+x^2/144)|)+C#

Jul 16, 2018

# -1/2xsqrt(x^2+144)+72ln|(sqrt(x^2+144)+x)/12|+C#.

Explanation:

Let, #I=int-(x^2/(sqrt(x^2+144)))dx=-intx^2/sqrt(x^2+144)dx#.

We subst. #x=12tany." Then, "dx=12sec^2ydy#.

#:. I=-int{(144tan^2y)(12sec^2y)}/sqrt(144tan^2y+144)dy#,

#=-144intsecytan^2ydy#,

#=-144int(tany)(secytanydy)#,

#=-144intsqrt(sec^2y-1)(secytanydy)#.

So, letting, #secy=t, &," so, "secytanydy=dt#, we have,

#I=-144intsqrt(t^2-1)dt#,

#=-144*1/2{tsqrt(t^2-1)-ln|t+sqrt(t^2-1)|}#,

#=-72{secytany-ln|secy+tany||#,

#=-72{tanysqrt(tan^2y+1)-ln|sqrt(tan^2y+1)+tany|}#,

#=-72{x/12*sqrt(x^2/144+1)-ln|sqrt(x^2/144+1)+x/12|}#.

#=-72{(xsqrt(x^2+144))/144-ln|(sqrt(x^2+144)+x)/12|}#.

#rArr I=-1/2xsqrt(x^2+144)+72ln|(sqrt(x^2+144)+x)/12|+C#.

N.B.:- The above Soln. has been worked out using trigo.**

substn. as it was so expected. But the same can be dealt

with without using any substn. as shown in the Second Soln.

What I had in my mind as the second soln., Respected Narad T.

has solved it exactly in the same way, so, I think, there is

no need for my second soln.

However, I prefer to use the following Standard Integrals :

#int1/sqrt(x^2+a^2)dx=ln|(x+sqrt(x^2+a^2))|+c_1, and, #

#intsqrt(x^2+a^2)dx=1/2{xsqrt(x^2+a^2)+a^2ln|(x+sqrt(x^2+a^2))|}+c_2#.