How do you integrate int -x^2/sqrt(144+x^2)dxx2144+x2dx using trigonometric substitution?

3 Answers
Jul 16, 2018

=>I=72ln|x+sqrt(x^2+144)|-x/2sqrt(x^2+144)+CI=72lnx+x2+144x2x2+144+C

Explanation:

Here ,

I=-intx^2/sqrt(x^2+144)dxI=x2x2+144dx

Subst. color(blue)(x=12tanu=>dx=12sec^2udu and x=12tanudx=12sec2uduand

color(blue)(tanu=x/12=>secu=sqrt(1+tan^2u)=sqrt(1+x^2/144)tanu=x12secu=1+tan2u=1+x2144

=>color(blue)(secu=sqrt(144+x^2)/12secu=144+x212

So,

I=-int(144tan^2u)/sqrt(144tan^2u+144)*12sec^2uduI=144tan2u144tan2u+14412sec2udu

=>I=-144int(tan^2u)/(12secu)12sec^2uduI=144tan2u12secu12sec2udu

=>I=-144intsecutan^2uduI=144secutan2udu

=>I=-144*I_1----to(A)I=144I1(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.