How do you evaluate the integral $\int \sqrt{x^{2} - 1} d x$ $int sqrt(x^2-1)dx$ ?

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How do you evaluate the integral $\int \sqrt{x^{2} - 1} d x$ $int sqrt(x^2-1)dx$ ?

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Answer 1 · 2017-01-22T17:30:23

$\int \sqrt{x^{2} - 1} . d x = \frac{1}{2} x \sqrt{x^{2} - 1} - \frac{1}{2} ln ∣ ∣ x + \sqrt{x^{2} - 1} ∣ ∣ + C$ $intsqrt(x^2-1)color(white).dx=1/2xsqrt(x^2-1)-1/2lnabs(x+sqrt(x^2-1))+C$

Explanation:

$I = \int \sqrt{x^{2} - 1} . d x$ $I=intsqrt(x^2-1)color(white).dx$

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

$I = \int \sqrt{{sec}^{2} θ - 1} (sec θ tan θ . d θ)$ $I=intsqrt(sec^2theta-1)(secthetatanthetacolor(white).d theta)$

Note that from the identity ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ we see that $tan θ = \sqrt{{sec}^{2} θ - 1}$ $tantheta=sqrt(sec^2theta-1)$ :

$I = \int tan (sec θ tan θ) d θ$ $I=inttan(secthetatantheta)d theta$

$I = \int sec θ {tan}^{2} θ . d θ$ $I=intsecthetatan^2thetacolor(white).d theta$

Let ${tan}^{2} θ = {sec}^{2} θ - 1$ $tan^2theta=sec^2theta-1$ :

$I = \int sec θ ({sec}^{2} θ - 1) . d θ$ $I=intsectheta(sec^2theta-1)color(white).d theta$

$I = \int {sec}^{3} θ . d θ - \int sec θ . d θ$ $I=intsec^3thetacolor(white).d theta-intsecthetacolor(white).d theta$

The integral of $sec θ$ $sectheta$ is common:

$I = \int {sec}^{3} θ . d θ - ln | sec θ + tan θ |$ $I=intsec^3thetacolor(white).d theta-lnabs(sectheta+tantheta)$

Let $J = \int {sec}^{3} θ . d θ$ $J=intsec^3thetacolor(white).d theta$ .

$J = \int sec θ ({sec}^{2} θ) d θ$ $J=intsectheta(sec^2theta)d theta$

This can be tackled using integration by parts. 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:

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

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

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

Returning to the original 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 $tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)$ .

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

Answer 2 · 2017-01-22T18:44:11

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

Explanation:

Let $I=intsqrt(x^2-1)dx=int(sqrt(x^2-1))(1)dx$

We will use the following Rule of Integration by Parts (IBP) :

$IBP : intuvdx=uintvdx-int{(du)/dxintvdx)}dx$

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,

$I=xsqrt(x^2-1)-int{(x/sqrt(x^2-1))(x)}dx$

$=xsqrt(x^2-1)-intx^2/sqrt(x^2-1)dx$

$=xsqrt(x^2-1)-int{(x^2-1)+1}/sqrt(x^2-1)dx$

$=xsqrt(x^2-1)-int{(x^2-1)/sqrt(x^2-1)+1/sqrt(x^2-1)}dx$

$=xsqrt(x^2-1)-intsqrt(x^2-1)dx-int1/sqrt(x^2-1)dx, i.e.,$

$I=xsqrt(x^2-1)-I-ln|x+sqrt(x^2-1)|$

$rArr I+I=2I=xsqrt(x^2-1)-ln|x+sqrt(x^2-1)|," and, therefore,"$

$I=x/2sqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+C$ .

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