How do you integrate $\int \frac{x^{3}}{\sqrt{64 + x^{2}}}$ $int x^3/sqrt(64+x^2)$ by trigonometric substitution?

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How do you integrate $\int \frac{x^{3}}{\sqrt{64 + x^{2}}}$ $int x^3/sqrt(64+x^2)$ by trigonometric substitution?

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Answer 1 · 2017-06-23T20:26:57

$I = \frac{64}{3} {(x^{2} + 64)}^{\frac{3}{2}} - 64 \sqrt{x^{2} + 64} + C$ $I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C$

Explanation:

Use the substitution $x = 8 tan θ$ $x= 8tantheta$ . Then $d x = 8 {sec}^{2} θ d θ$ $dx= 8sec^2theta d theta$ .

$I = \int \frac{{(8 tan θ)}^{3}}{\sqrt{64 + {(8 tan θ)}^{2}}} \cdot 8 {sec}^{2} θ d θ$ $I = int (8tantheta)^3/sqrt(64 + (8tantheta)^2) * 8sec^2theta d theta$

$I = \int \frac{512 {tan}^{3} θ}{\sqrt{64 + 64 {tan}^{2} θ}} \cdot 8 {sec}^{2} θ d θ$ $I = int (512tan^3theta)/sqrt(64 + 64tan^2theta) * 8sec^2theta d theta$

$I = \int \frac{512 {tan}^{3} θ}{\sqrt{64 (1 + {tan}^{2} θ)}} \cdot 8 {sec}^{2} θ d θ$ $I = int(512tan^3theta)/sqrt(64(1 + tan^2theta)) * 8sec^2theta d theta$

Now use $1 + {tan}^{2} θ = {sec}^{2} θ$ $1 + tan^2theta = sec^2theta$ .

$I = \int \frac{512 {tan}^{3} θ}{\sqrt{64 {sec}^{2} θ}} \cdot 8 {sec}^{2} θ d θ$ $I = int(512tan^3theta)/sqrt(64sec^2theta) * 8sec^2theta d theta$

$I = \int \frac{512 {tan}^{3} θ}{8 sec θ} \cdot 8 {sec}^{2} θ d θ$ $I = int(512tan^3theta)/(8sectheta) * 8sec^2theta d theta$

$I = \int 512 sec θ {tan}^{3} θ d θ$ $I = int512sec thetatan^3theta d theta$

$I = \int 512 sec θ tan θ ({tan}^{2} θ) d θ$ $I =int512secthetatantheta(tan^2theta) d theta$

We now use ${tan}^{2} θ = {sec}^{2} θ - 1$ $tan^2theta = sec^2theta -1$ .

$I = \int 512 sec θ tan θ ({sec}^{2} θ - 1) d θ$ $I = int512secthetatantheta(sec^2theta - 1) d theta$

Now let $u = sec θ$ $u = sectheta$ . Then $d u = sec θ tan θ d θ$ $du = secthetatantheta d theta$ and $d θ = \frac{d u}{sec θ tan θ}$ $d theta = (du)/(secthetatantheta)$ .

$I = \int 512 sec θ tan θ \cdot (u^{2} - 1) \frac{d u}{sec θ tan θ}$ $I = int512secthetatantheta * (u^2- 1) (du)/(secthetatantheta)$

$I = \int (512 u^{2} - 512) d u$ $I= int(512u^2 - 512)du$

$I = \frac{512}{3} u^{3} - 512 u + C$ $I = 512/3u^3 - 512u + C$

$I = \frac{512}{3} {sec}^{3} θ - 512 sec θ$ $I = 512/3sec^3theta - 512sectheta$

From our initial substitution, we know that $tan θ = \frac{x}{8}$ $tantheta = x/8$ . Then $sec θ = \frac{\sqrt{x^{2} + 64}}{8}$ $sectheta = sqrt(x^2 + 64)/8$

$I = \frac{64}{3} {(x^{2} + 64)}^{\frac{3}{2}} - 64 \sqrt{x^{2} + 64} + C$ $I = 64/3(x^2 + 64)^(3/2) - 64sqrt(x^2 + 64) + C$

Hopefully this helps!

Answer 2 · 2017-06-24T09:57:34

$\int \frac{x^{3}}{\sqrt{64 + x^{2}}} d x = \frac{1}{3} \sqrt{64 + x^{2}} (x^{2} - 128) + C$ $int (x^3)/(sqrt{64+x^2}) "d"x= 1/3sqrt{64+x^2}(x^2-128) + C$ .

Explanation:

I know you specified trigonometric substitution but I don't see why you'd use on in this case as a more obvious one stands out to me, because we have an $x^{2}$ $x^2$ with an $x^{3}$ $x^3$ on the numerator.

Someone's already submitted the trigonometric substitution method so thought I may as well share.

$I = \int \frac{x^{3}}{\sqrt{64 + x^{2}}} d x$ $I = int (x^3)/(sqrt{64+x^2})"d"x$ .

Let $u = x^{2}$ $u=x^2$ , $\frac{d u}{d x} = 2 x$ $("d"u)/("d"x) = 2x$ , $d x = \frac{1}{2 x} d u$ $"d"x = 1/(2x) "d"u$ .

Then,

$I = \frac{1}{2} \int \frac{x^{2}}{\sqrt{64 + x^{2}}} d u$ $I = 1/2 int x^2/(sqrt{64+x^2}) "d"u$ ,

$= \frac{1}{2} \int u {(64 + u)}^{- \frac{1}{2}} d u$ $= 1/2 int u(64+u)^(-1/2) "d"u$ .

At this point, I used integration by parts. Let $s = u$ $s = u$ , and $d t = {(64 + u)}^{- \frac{1}{2}} d u$ $dt = (64 + u)^(-1/2)du$ . Then, for $\int s d t = s t - \int t d s$ $int sdt = st - int tds$ :

$I = \frac{1}{2} ([2 u {(64 + u)}^{\frac{1}{2}}] - 2 \int {(64 + u)}^{\frac{1}{2}} d u)$ $I = 1/2 ( [2u(64+u)^(1/2)]-2int(64+u)^(1/2) "d"u)$

$I = u {(64 + u)}^{\frac{1}{2}} - \frac{2}{3} {(64 + u)}^{\frac{3}{2}} + C$ $I = u(64+u)^(1/2) - 2/3(64+u)^(3/2) + C$

Then, I factored out $\frac{1}{3} \sqrt{64 + x^{2}}$ $1/3 sqrt(64 + x^2)$ and substituted $u = x^{2}$ $u = x^2$ back in:

$I = \frac{1}{3} \sqrt{64 + x^{2}} (3 x^{2} - 2 (64 + x^{2})) + C$ $I = 1/3sqrt{64+x^2}(3x^2-2(64+x^2)) + C$

$= \frac{1}{3} \sqrt{64 + x^{2}} (x^{2} - 128) + C$ $= color(blue)(1/3sqrt{64+x^2}(x^2-128) + C)$ .

Answer 3 · 2017-06-27T14:10:57

$\frac{1}{3} \sqrt{x^{2} + 64} (x^{2} - 128) + C .$ $1/3sqrt(x^2+64)(x^2-128)+C.$

Explanation:

Here is another way to solve the Problem, without using

substitution.

$I = \int \frac{x^{3}}{\sqrt{x^{2} + 64}} d x,$ $I=intx^3/sqrt(x^2+64)dx,$

$= \int \frac{x^{2} \cdot x}{\sqrt{x^{2} + 64}} d x,$ $=int(x^2*x)/sqrt(x^2+64)dx,$

$= \int \frac{((x^{2} + 64) - 64) x}{\sqrt{x^{2} + 64}} d x,$ $=int{((x^2+64)-64)x}/sqrt(x^2+64)dx,$

$= \int (\frac{x^{2} + 64}{\sqrt{x^{2} + 64}} - \frac{64}{\sqrt{x^{2} + 64}}) x d x,$ $=int((x^2+64)/sqrt(x^2+64)-64/sqrt(x^2+64))xdx,$

$= \int {{(x^{2} + 64)}^{\frac{1}{2}} - 64 {(x^{2} + 64)}^{- \frac{1}{2}}} {\frac{1}{2} \frac{d}{d x} (x^{2} + 64)} d x,$ $=int{(x^2+64)^(1/2)-64(x^2+64)^(-1/2)}{1/2d/dx(x^2+64)}dx,$

$= \int {(x^{2} + 64)}^{\frac{1}{2}} {\frac{1}{2} \frac{d}{d x} (x^{2} + 64)} d x$ $=int(x^2+64)^(1/2){1/2d/dx(x^2+64)}dx$
$- 64 \int {(x^{2} + 64)}^{- \frac{1}{2}} {\frac{1}{2} \frac{d}{d x} (x^{2} + 64)} d x,$ $-64int(x^2+64)^(-1/2){1/2d/dx(x^2+64)}dx,$

$= \frac{1}{2} \int {(x^{2} + 64)}^{\frac{1}{2}} (\frac{d}{d x} (x^{2} + 64)) d x$ $=1/2int(x^2+64)^(1/2)(d/dx(x^2+64))dx$
$- 32 \int {(x^{2} + 64)}^{- \frac{1}{2}} (\frac{d}{d x} (x^{2} + 64)) d x,$ $-32int(x^2+64)^(-1/2)(d/dx(x^2+64))dx,$

$= \frac{1}{2} \frac{{(x^{2} + 64)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} - 32 \frac{{(x^{2} + 64)}^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1},$ $=1/2(x^2+64)^(1/2+1)/(1/2+1)-32(x^2+64)^(-1/2+1)/(-1/2+1),$

$= \frac{1}{3} {(x^{2} + 64)}^{\frac{3}{2}} - 64 {(x^{2} + 64)}^{\frac{1}{2}},$ $=1/3(x^2+64)^(3/2)-64(x^2+64)^(1/2),$

$= \frac{1}{3} \sqrt{x^{2} + 64} {(x^{2} + 64) - 192)},$ $=1/3sqrt(x^2+64){(x^2+64)-192)},$

$= \frac{1}{3} \sqrt{x^{2} + 64} (x^{2} - 128) + C .$ $=1/3sqrt(x^2+64)(x^2-128)+C.$

N.B.: The Final Integrals were obtained using the Rule,

$int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c, n ne-1.$

Enjoy Maths.!

Answer 4 · 2017-07-09T20:46:58

After using $x=8sinhu$ and $dx=8coshu$ transforms, this integral became,

$int(8sinhu)^3*(8coshu)/(8coshu)*du$

$=int512(sinhu)^3*du$

$=int512(sinhu)^2*sinhu*du$

$=int512((coshu)^2-1)*sinhu*du$

$=int512(coshu)^2*sinhu*du-int512sinhu*du$

$=512/3*(coshu)^3-512coshu+C$

After using $sinhu=x/8$ and $coshu=(Sqrt(x^2+64))/8$ inverse transforms, solution of this integral became,

= $int(x^3)/(Sqrt(x^2+64))*dx$

= $=1/3*(x^2+64)^(3/2)-64Sqrt(x^2+64)+C$

= $=1/3*(x^2-128)*Sqrt(x^2+64)+C$

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