How do you integrate int sqrt(x^2-25) dx using trigonometric substitution?

1 Answer
May 25, 2018

color(green)[int sqrt(x^2-25)*dx=-(25*ln(abs(sqrt(x^2-25)+x))-x*sqrt(x^2-25))/2]

Explanation:

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