Question

Integrate `x^3*√(16-x^2)` by using Trigonometric Substitutions (only)???

Answer 1

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

Explanation:

Here,

`I=intx^3sqrt(16-x^2)dx`

Let,

`x=4sinu=>dx=4cosudu , and`

`sinu=x/4`
`=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16)=sqrt(16-x^2)/4...to(1)`

So,

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

`=int64sin^3u(4cosu)4cosudu`

`=1024intsin^3ucos^2udu`

`=1024intsin^2ucos^2usinudu`

`=1024int(1-cos^2u)cos^2usinudu`

`=1024int[cos^2usinu-cos^4usinu]du`

`=1024[-int(cosu)^2(-sinu)du+int(cosu)^4(-sinu)du]`

`=1024[-int(cosu)^2(d/(dx)(cosu))du+int(cosu)^4(d/(dx)(cosu)du]`

`=1024[-cos^3u/3+cos^5u/5]+c`

`=1024[(cosu)^5/5-(cosu)^3/3]+c`

From `(1)`,we have `cosu=sqrt(16-x^2)/4`

#=4^5[(sqrt(16-x^2))^5/(4^5xx5)-(sqrt(16- x^2))^3/(4^3xx3)]+c#

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

Answer 2

`I =1/5(16 - x^2)^(5/2)- 16/3(16 - x^2)^(3/2)+C`

Explanation:

Let `x =4sintheta`. Then `dx= 4costheta d theta`.

`I = int (4sintheta)^3sqrt(16- (4sintheta)^2) d theta * 4costhetad theta`

`I =int 64sin^3theta (4costheta)(4costheta)d theta`

`I = 1024int sin^3thetacos^2theta d theta`

`I = 1024intsin^2thetasinthetacos^2theta d theta`

`I = 1024int (1- cos^2theta)cos^2thetasintheta`

`I = 1024int cos^2thetasintheta - 1024cos^4thetasinthetad theta`

`I = 1024int cos^2thetasintheta d theta - 1024int cos^4theta sin theta d theta`

We now let `u = costheta`. Then `du = -sinthetad theta` and `d theta= (du)/(-sintheta)`.

`I =1024int u^2 du - 1024int u^4 du`

`I = -1024(1/3u^3) +1024(1/5u^5) + C`

`I = -1024/3u^3 +1024/5u^5 + C`

`I = -1024/3cos^3theta + 1024/5cos^5theta + C`

From our initial substitution we see that `x/4 = sintheta`. Therefore, `sqrt(16 - x^2)/4 = costheta`.

`I = -1024/3(sqrt(16 -x^2)/4)^3 + 1024/5(sqrt(16 - x^2)/4)^5 + C`

`I = -16/3(16 - x^2)^(3/2) +1/5(16 - x^2)^(5/2) + C`

Hopefully this helps!