How do you evaluate the integral #int sqrt(x^2-1)dx#?

Question

Ezra Adams · Accepted Answer

#intsqrt(x^2-1)color(white).dx=1/2xsqrt(x^2-1)-1/2lnabs(x+sqrt(x^2-1))+C#

#I=intsqrt(x^2-1)color(white).dx#

Use the trigonometric substitution #x=sectheta#. Differentiating this shows that #dx=secthetatanthetacolor(white).d theta#. Substituting both of these gives:

#I=intsqrt(sec^2theta-1)(secthetatanthetacolor(white).d theta)#

Note that from the identity #tan^2theta+1=sec^2theta# we see that #tantheta=sqrt(sec^2theta-1)#:

#I=inttan(secthetatantheta)d theta#

#I=intsecthetatan^2thetacolor(white).d theta#

Let #tan^2theta=sec^2theta-1#:

#I=intsectheta(sec^2theta-1)color(white).d theta#

#I=intsec^3thetacolor(white).d theta-intsecthetacolor(white).d theta#

The integral of #sectheta# is common:

#I=intsec^3thetacolor(white).d theta-lnabs(sectheta+tantheta)#

Let #J=intsec^3thetacolor(white).d theta#.

#J=intsectheta(sec^2theta)d theta#

Integration by parts can be used to address this. Let:

#{(u=sectheta" "=>" "du=secthetatanthetacolor(white).d theta),(dv=sec^2thetacolor(white).d theta" "=>" "v=tantheta):}#

Then:

#J=secthetatantheta-intsecthetatan^2thetacolor(white).d theta#

Again let #tan^2theta=sec^2theta-1#:

#J=secthetatantheta-intsectheta(sec^2theta-1)d theta#

#J=secthetatantheta-intsec^3thetacolor(white).d theta+intsecthetacolor(white).d theta#

Note that the original integral #J# is included again, and we know the integral of secant:

#J=secthetatantheta-J+lnabs(sectheta+tantheta)#

#2J=secthetatantheta+lnabs(sectheta+tantheta)#

#J=1/2secthetatantheta+1/2lnabs(sectheta+tantheta)#

Going back to the initial integral:

#I=J-lnabs(sectheta+tantheta)#

#I=(1/2secthetatantheta+1/2lnabs(sectheta+tantheta))-lnabs(sectheta+tantheta)#

#I=1/2secthetatantheta-1/2lnabs(sectheta+tantheta)#

Recall our original substitution #x=sectheta#. This implies that #tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)#.

#I=1/2xsqrt(x^2-1)-1/2lnabs(x+sqrt(x^2-1))+C#

Ezekiel Alexander · Answer

#x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C#.

Let #I=intsqrt(x^2-1)dx=int(sqrt(x^2-1))(1)dx# The following Integration by Parts (IBP) Rule will be applied: # IBP : intuvdx=uintvdx-int{(du)/dxintvdx)}dx# Taking #u=sqrt(x^2-1) rArr (du)/dx=1/(2sqrt(x^2-1))d/dx(x^2-1), i.e.,# #(du)/dx=x/sqrt(x^2-1)# #v=1 rArr intvdx=x# Therefore, #I=xsqrt(x^2-1)-int{(x/sqrt(x^2-1))(x)}dx# #=xsqrt(x^2-1)-intx^2/sqrt(x^2-1)dx# #=xsqrt(x^2-1)-int{(x^2-1)+1}/sqrt(x^2-1)dx# #=xsqrt(x^2-1)-int{(x^2-1)/sqrt(x^2-1)+1/sqrt(x^2-1)}dx# #=xsqrt(x^2-1)-intsqrt(x^2-1)dx-int1/sqrt(x^2-1)dx, i.e.,# #I=xsqrt(x^2-1)-I-ln|x+sqrt(x^2-1)|# #rArr I+I=2I=xsqrt(x^2-1)-ln|x+sqrt(x^2-1)|," and, therefore,"# #I=x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C#. Share the Happiness, Math!

Emily Ammerman · Answer

undefined To evaluate the integral ∫√(x^2 - 1) dx, you can use trigonometric substitution. Let x = sec(θ), then dx = sec(θ)tan(θ) dθ. Substituting these into the integral, you get: ∫√(sec^2(θ) - 1) sec(θ)tan(θ) dθ. Simplifying, sec^2(θ) - 1 = tan^2(θ), so the integral becomes: ∫tan^2(θ)sec(θ)tan(θ) dθ. This simplifies to: ∫tan^3(θ) sec(θ) dθ. Use the reduction formula for integrating powers of tangent to solve this integral. After integrating, convert back to x using the original substitution.