The integral is int sqrt(x^2-25)*dx
Let suppose:
x=5sectheta
dx=5*sectheta*tantheta*d(theta)
sqrt(x^2-25)=sqrt[25(sec^2theta-1)]=5tantheta
the integral become after suppose:
int sqrt(x^2-25)*dx=int5tantheta*5sectheta*tantheta*d(theta)
25intsectheta*tan^2theta*d(theta)=25intsectheta*(sec^2theta-1)*d(theta)
25intsec^3theta-sectheta*d(theta)=25intsec^3theta-25intsectheta*d(theta)
Firstly let solve color(red)[intsec^3theta*d(theta) by using integral y parts:
color(red)(I=intsec^3(theta)d(theta)...to(1)
"Using "color(blue)"Integration by Parts"
int(u*v)dx=uintvdx-int(u'intvdx)d(theta)
Let u=sec(theta) and v=sec^2(theta)
=>u'=sec(theta)tan(theta) and intvd(theta)=tan(theta)
I=sec(theta) xxtan(theta)-intsec(theta)tantheta xxtan(theta) d(theta)
=secthetatan(theta)-intsec(theta)tan^2(theta)dx+c
=sec(theta)tan(theta)-intsec(theta)(sec^2(theta)-1)d(theta)+c
I=sec(theta)tan(theta)-color(red)(intsec^3(theta)dx)+intsec(theta)d(theta)+c
I=sec(theta)tan(theta)-color(red)(I)+intsec(theta)d(theta)+c...tofrom1
I+I=sec(theta)tan(theta)+intsec(theta)d(theta)+c
2I=sec(theta)tan(theta)+ln|sec(theta)+tan(theta)|+C
I=1/2sec(theta)tan(theta)+1/2ln|sec(theta)+tan(theta)|+C
secondly let find color(blue)[intsectheta*d(theta)
intsectheta*d(theta)=intsectheta*[sectheta+tantheta]/[sectheta+tantheta]*d(theta)
int(sec^2theta+sectheta*tantheta)/(sectheta+tantheta)*d(theta)=ln|sectheta+tantheta|+c
color(green)[25intsec^3theta-25intsectheta*d(theta)]=
25[1/2sec(theta)tan(theta)+1/2ln|sec(theta)+tan(theta)|]-25[ln|sectheta+tantheta|]+c
=(x*sqrt(x^2-25))/2-(25*ln(abs(sqrt(x^2-25)+x)))/2
After simplified it we will get:
=-(25*ln(abs(sqrt(x^2-25)+x))-x*sqrt(x^2-25))/2