How do you integrate $\int x^{3} \sqrt{16 - x^{2}}$ $int x^3sqrt(16-x^2)$ by trigonometric substitution?

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How do you integrate $\int x^{3} \sqrt{16 - x^{2}}$ $int x^3sqrt(16-x^2)$ by trigonometric substitution?

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Answer 1 · 2018-05-05T16:22:53

$\int x^{3} \sqrt{16 - x^{2}} d x = - \frac{16}{3} {(16 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} {(16 - x^{2})}^{\frac{5}{2}} + c$ $intx^3sqrt(16-x^2)dx=-16/3(16-x^2)^(3/2)+1/5(16-x^2)^(5/2)+c$

Explanation:

Let $x = 4 sin t$ $x=4sint$ , then $\sqrt{16 - x^{2}} = 4 cos t$ $sqrt(16-x^2)=4cost$ and $d x = 4 cos t d t$ $dx=4costdt$

and $\int x^{3} \sqrt{16 - x^{2}} d x$ $intx^3sqrt(16-x^2)dx$

= $\int 64 {sin}^{3} t \cdot 4 cos t \cdot 4 cos t d t$ $int64sin^3t*4cost*4costdt$

= $1024 \int {sin}^{3} t {cos}^{2} t d t$ $1024intsin^3tcos^2tdt$

= $1024 \int (1 - {cos}^{2} t) {cos}^{2} t sin t d t$ $1024int(1-cos^2t)cos^2tsintdt$

= $1024 \int ({cos}^{2} t - {cos}^{4} t) sin t d t$ $1024int(cos^2t-cos^4t)sintdt$

Let $cos t = u$ $cost=u$ then $d u = - sin t d t$ $du=-sintdt$ and our integral becomes

$- 1024 \int (u^{2} - u^{4}) d u$ $-1024int(u^2-u^4)du$

= $- 1024 (\frac{u^{3}}{3} - \frac{u^{5}}{5})$ $-1024(u^3/3-u^5/5)$

i.e. $1024 (- \frac{1}{3} {cos}^{3} t + \frac{1}{5} {cos}^{5} t)$ $1024(-1/3cos^3t+1/5cos^5t)$ (substituting $u = cos t$ $u=cost$ )

and as $cos t = \frac{1}{4} \sqrt{16 - x^{2}}$ $cost=1/4sqrt(16-x^2)$ , our integral becomes

$1024 (- \frac{1}{3} \cdot \frac{1}{64} {(16 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} \cdot \frac{1}{1024} {(16 - x^{2})}^{\frac{5}{2}}) + c$ $1024(-1/3*1/64(16-x^2)^(3/2)+1/5*1/1024(16-x^2)^(5/2))+c$

= $- \frac{16}{3} {(16 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} {(16 - x^{2})}^{\frac{5}{2}} + c$ $-16/3(16-x^2)^(3/2)+1/5(16-x^2)^(5/2)+c$

Answer 2 · 2018-05-05T16:34:41

Let, $\sqrt{16 - x^{2}} = t \Rightarrow 16 - x^{2} = t^{2} \Rightarrow x d x = - t d t$ $sqrt(16-x^2)=t=>16-x^2=t^2=>xdx=-tdt$
$I = \int (16 - t^{2}) t (- t) d t = \int (t^{4} - 16 t^{2}) d t = \frac{t^{5}}{5} - \frac{16 t^{3}}{3} + c$ $I=int(16-t^2)t(-t)dt=int(t^4-16t^2)dt=t^5/5-(16t^3)/3+c$
Substituting, $t = \sqrt{16 - x^{2}}$ $t=sqrt(16-x^2)$
$I = \frac{{(\sqrt{16 - x^{2}})}^{5}}{5} - \frac{16 {(\sqrt{16 - x^{2}})}^{3}}{3} + c$ $I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+c$

Explanation:

Here,

$I = \int x^{3} \sqrt{16 - x^{2}} d x$ $I=intx^3sqrt(16-x^2)dx$

Let, $x = 4 sin u \Rightarrow d x = 4 cos u d u$ $x=4sinu=>dx=4cosudu$

$and sin u = \frac{x}{4} \Rightarrow cos u = \sqrt{1 - {sin}^{2} u} = \sqrt{1 - \frac{x^{2}}{16}}$ $and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16$

$=>cosu=sqrt(16-x^2)/4...to(A)$

$I=int64sin^3usqrt(16-16sin^2u)xx4cosudu$

$=256intsin^3uxx4cosuxxcosudu$

$I=1024intsin^3ucos^2udu$

$I=4^5int(1-cos^2u)cos^2uxxsinudu$

$=4^5[intcos^2usinudu-intcos^4usinudu]$

$=4^5[-int(cosu)^2(-sinu)du+int(cosu)^4(-sinu)du]$

$=4^5[-(cosu)^3/3+(cosu)^5/5]+c$

$=4^5[(cosu)^5/5-(cosu)^3/3]+c...to$ from $(A)$

$=4^5[(sqrt(16-x^2))^5/(5xx4^5)-(sqrt(16-x^2))^3/(3xx4^3)]+c$

$=(sqrt(16-x^2))^5/5-4^2xx(sqrt(16-x^2))^3/3+c$

$I=(sqrt(16-x^2))^5/5-(16(sqrt(16-x^2))^3)/3+c$

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