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

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#

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!