How do you solve $\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x$ $intx^2/(sqrt(x^2-9))dx$ using trig substitution?

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How do you solve $\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x$ $intx^2/(sqrt(x^2-9))dx$ using trig substitution?

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Answer 1 · 2018-04-18T12:07:25

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = \frac{x \sqrt{x^{2} - 9}}{2} + \frac{9}{2} ln ∣ ∣ x + \sqrt{x^{2} - 9} ∣ ∣ + C$ $int x^2/sqrt(x^2-9)dx = (xsqrt(x^2-9))/2 +9/2ln abs (x +sqrt(x^2-9)) +C$

Explanation:

The integrand is defined for $x \in (- \infty, - 3) \cup (3, + \infty)$ $x in(-oo,-3) uu (3,+oo)$ . Let us focus for the moment on $x \in (3, + \infty)$ $x in (3,+oo)$ and substitute:

$x = 3 sec t$ $x = 3sect$

with $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$

so that:

$d x = 3 sec t tan t d t$ $dx = 3sect tant dt$

and:

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = \int \frac{{(3 sec t)}^{2} (3 sec t tan t) d t}{\sqrt{{(3 sec t)}^{2} - 9}}$ $int x^2/sqrt(x^2-9)dx = int ((3sect)^2(3sect tant) dt)/sqrt((3sect)^2-9)$

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = 27 \int \frac{{sec}^{3} t tan t d t}{\sqrt{9 {sec}^{2} t - 9}}$ $int x^2/sqrt(x^2-9)dx = 27 int (sec^3t tant dt)/sqrt(9sec^2t-9)$

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = 9 \int \frac{{sec}^{3} t tan t d t}{\sqrt{{sec}^{2} t - 1}}$ $int x^2/sqrt(x^2-9)dx = 9 int (sec^3t tant dt)/sqrt(sec^2t-1)$

Use now the trigonometric identity:

${sec}^{2} t - 1 = {tan}^{2} t$ $sec^2t -1 = tan^2t$

and considering that for $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$ the tangent is positive:

$\sqrt{{sec}^{2} t - 1} = tan t$ $sqrt(sec^2t -1) = tant$

so:

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = 9 \int \frac{{sec}^{3} t tan t d t}{tan t}$ $int x^2/sqrt(x^2-9)dx = 9 int (sec^3t tant dt)/tant$

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = 9 \int {sec}^{3} t d t$ $int x^2/sqrt(x^2-9)dx = 9 int sec^3tdt$

Now write ${sec}^{3} t$ $sec^3t$ as $sec t \times {sec}^{2} t$ $sect xx sec^2t$ , and as $\frac{d}{d t} tan t = {sec}^{2} t$ $d/dt tant = sec^2t$ we can integrate by parts:

$\int {sec}^{3} t d t = \int sec t d (tan t)$ $int sec^3tdt = int sect d(tant)$

$\int {sec}^{3} t d t = sec t tan t - \int tan t d (sec t)$ $int sec^3tdt = sect tant - int tant d(sect )$

$\int {sec}^{3} t d t = sec t tan t - \int {tan}^{2} t sec t d t$ $int sec^3tdt = sect tant - int tan^2tsect dt$

using the same identity again:

$\int {sec}^{3} t d t = sec t tan t - \int ({sec}^{2} t - 1) sec t d t$ $int sec^3tdt = sect tant - int (sec^2t-1)sect dt$

and based in the linearity of the integral:

$\int {sec}^{3} t d t = sec t tan t - \int {sec}^{3} t + \int sec t d t$ $int sec^3tdt = sect tant - int sec^3t+ int sect dt$

The integral now appears on both sides of the equation and we can solve for it:

$2 \int {sec}^{3} t d t = sec t tan t + \int sec t d t$ $2int sec^3tdt = sect tant + int sect dt$

and finally :

$\int {sec}^{3} t d t = \frac{sec t tan t}{2} + \frac{1}{2} ln | sec t + tan t | + C$ $int sec^3tdt = (sect tant)/2 + 1/2 ln abs (sect+tant)+C$

Undo now the substitution:

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = \frac{9 sec t tan t}{2} + \frac{9}{2} ln | sec t + tan t | + C$ $int x^2/sqrt(x^2-9)dx = (9sect tant)/2 + 9/2 ln abs (sect+tant)+C$

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = \frac{3 x \sqrt{{(\frac{x}{3})}^{2} - 1}}{2} + \frac{9}{2} ln ∣ ∣ ∣ \frac{x}{3} + \sqrt{{(\frac{x}{3})}^{2} - 1} ∣ ∣ ∣ + C$ $int x^2/sqrt(x^2-9)dx = (3xsqrt((x/3)^2-1) )/2 + 9/2 ln abs (x/3+sqrt((x/3)^2-1))+C$

$\int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x = \frac{x \sqrt{x^{2} - 9}}{2} + \frac{9}{2} ln ∣ ∣ x + \sqrt{x^{2} - 9} ∣ ∣ + C$ $int x^2/sqrt(x^2-9)dx = (xsqrt(x^2-9))/2 +9/2ln abs (x +sqrt(x^2-9)) +C$

By direct differentiation we can see that the solution is valid also for $x \in (- \infty, - 3)$ $x in (-oo,-3)$ .

Answer 2 · 2018-04-19T05:27:33

$\frac{1}{2} {x \sqrt{x^{2} - 9} + 9 ln ∣ ∣ (x + \sqrt{x^{2} - 9}) ∣ ∣} + C$ $1/2{xsqrt(x^2-9)+9ln|(x+sqrt(x^2-9))|}+C$ .

Explanation:

Let us solve the Problem without using substn.

$I = \int \frac{x^{2}}{\sqrt{x^{2} - 9}} d x$ $I=intx^2/sqrt(x^2-9)dx$ .

$:. I=int{(x^2-9)+9}/sqrt(x^2-9)dx$ ,

$=int{(x^2-9)/sqrt(x^2-9)+9/sqrt(x^2-9)}dx$ ,

$=intsqrt(x^2-9)dx+9int1/sqrt(x^2-9)dx$ ,

$={x/2sqrt(x^2-9)-9/2ln|(x+sqrt(x^2-9))|}+9ln|(x+sqrt(x^2-9))|$ ,

$rArr I=1/2{xsqrt(x^2-9)+9ln|(x+sqrt(x^2-9))|}+C$ .

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