I would let:
x = 4sinthetax=4sinθ
dx = 4costhetad thetadx=4cosθdθ
sqrt(16 - x^2) = 4costheta√16−x2=4cosθ
x^3 = 64sin^3thetax3=64sin3θ
=>1024int sin^3thetacos^2thetad theta⇒1024∫sin3θcos2θdθ
=1024int sin^2thetasinthetacos^2thetad theta=1024∫sin2θsinθcos2θdθ
=1024int (1-cos^2theta)cos^2thetasinthetad theta=1024∫(1−cos2θ)cos2θsinθdθ
At this point you can let:
u = costhetau=cosθ
du = -sinthetad thetadu=−sinθdθ
=> 1024int (cos^2theta - 1)cos^2theta(-sintheta)d theta⇒1024∫(cos2θ−1)cos2θ(−sinθ)dθ
= 1024int (u^2 - 1)u^2du=1024∫(u2−1)u2du
= 1024int u^4 - u^2du=1024∫u4−u2du
= 1024(u^5/5 - u^3/3)=1024(u55−u33)
= 1024(cos^5theta/5 - cos^3theta/3)=1024(cos5θ5−cos3θ3)
Since costheta = sqrt(16-x^2)/4cosθ=√16−x24:
= 1024((sqrt(16-x^2)/4)^5/5 - (sqrt(16-x^2)/4)^3/3)=1024⎛⎜
⎜⎝(√16−x24)55−(√16−x24)33⎞⎟
⎟⎠
= 1024(((16-x^2)^(5/2)/1024)/5 - ((16-x^2)^(3/2)/64)/3)=1024⎛⎜
⎜⎝(16−x2)5210245−(16−x2)32643⎞⎟
⎟⎠
= 1024((16-x^2)^(5/2)/(5*1024) - (16-x^2)^(3/2)/(3*64))=1024⎛⎜⎝(16−x2)525⋅1024−(16−x2)323⋅64⎞⎟⎠
= cancel(1024)((16-x^2)^(5/2)/(5*cancel(1024)) - (16-x^2)^(3/2)/(3/16*cancel(1024)))
= color(green)(1/5(16-x^2)^("5/2") - 16/3(16-x^2)^("3/2") + C)
Normally here would be okay, but we can go a bit further.
= (16-x^2)^(3/2)(1/5(16-x^2) - 16/3)
= (16-x^2)^(3/2)(3/15(16-x^2) - 80/15)
= 1/15(16-x^2)^(3/2)(3(16-x^2) - 80)
= 1/15(16-x^2)^(3/2)(48-3x^2 - 80)
= 1/15(16-x^2)^(3/2)(-32-3x^2)
= color(blue)(-1/15(16-x^2)^("3/2")(3x^2 + 32) + C)
