Question

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

Answer 1

`=>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`

Answer 2

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`

Answer 3

`-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`.