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...to#from1
#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#
