How do you integrate #intx^3sqrt(16 - x^2) dx#?

Answer 1
I would let: #x = 4sintheta# #dx = 4costhetad theta# #sqrt(16 - x^2) = 4costheta# #x^3 = 64sin^3theta#
#=>1024int sin^3thetacos^2thetad theta#
#=1024int sin^2thetasinthetacos^2thetad theta#
#=1024int (1-cos^2theta)cos^2thetasinthetad theta#
At this point you can let: #u = costheta# #du = -sinthetad theta#
#=> 1024int (cos^2theta - 1)cos^2theta(-sintheta)d theta#
#= 1024int (u^2 - 1)u^2du#
#= 1024int u^4 - u^2du#
#= 1024(u^5/5 - u^3/3)#
#= 1024(cos^5theta/5 - cos^3theta/3)#
Since #costheta = sqrt(16-x^2)/4#:
#= 1024((sqrt(16-x^2)/4)^5/5 - (sqrt(16-x^2)/4)^3/3)#
#= 1024(((16-x^2)^(5/2)/1024)/5 - ((16-x^2)^(3/2)/64)/3)#
#= 1024((16-x^2)^(5/2)/(5*1024) - (16-x^2)^(3/2)/(3*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)#
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Answer 2

To integrate ( \int x^3 \sqrt{16 - x^2} , dx ), you can use the substitution method. Let ( u = 16 - x^2 ). Then ( du = -2x , dx ). This implies ( x , dx = -\frac{1}{2} du ).

Substituting these into the integral, you get:

[ \int x^3 \sqrt{16 - x^2} , dx = \int -\frac{1}{2} \left(16 - u\right)^{\frac{1}{2}} \cdot u , du ]

Now, you can use integration by parts. Let ( dv = u , du ) and ( v = \frac{2}{3} (16 - u)^{\frac{3}{2}} ).

[ \int -\frac{1}{2} \left(16 - u\right)^{\frac{1}{2}} \cdot u , du = -\frac{1}{2} \cdot \frac{2}{3} (16 - u)^{\frac{3}{2}} \cdot u + \frac{1}{2} \int \frac{2}{3} (16 - u)^{\frac{3}{2}} , du ]

Now, integrate the remaining integral:

[ \frac{1}{2} \int \frac{2}{3} (16 - u)^{\frac{3}{2}} , du = -\frac{1}{3} (16 - u)^{\frac{3}{2}} + C ]

Combine the results:

[ \int x^3 \sqrt{16 - x^2} , dx = -\frac{1}{3} (16 - u)^{\frac{3}{2}} - \frac{1}{2} \cdot \frac{2}{3} (16 - u)^{\frac{3}{2}} \cdot u + C ]

Replace ( u ) with ( 16 - x^2 ) to get the final result:

[ \int x^3 \sqrt{16 - x^2} , dx = -\frac{1}{3} (16 - 16 + x^2)^{\frac{3}{2}} - \frac{1}{3} (16 - x^2)^{\frac{3}{2}} + C ]

[ \int x^3 \sqrt{16 - x^2} , dx = -\frac{1}{3} (x^2)^{\frac{3}{2}} - \frac{1}{3} (16 - x^2)^{\frac{3}{2}} + C ]

[ \int x^3 \sqrt{16 - x^2} , dx = -\frac{1}{3} x^3 - \frac{1}{3} (16 - x^2)^{\frac{3}{2}} + C ]

This is the final result after integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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