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What is the integral #int_0^(2sqrt(3)) x^3/sqrt(16 - x^2) dx#?

Question

What is the integral #int_0^(2sqrt(3)) x^3/sqrt(16 - x^2) dx#?
What is the integral #int_0^(2sqrt(3)) x^3/sqrt(16 - x^2) dx#?

Answers

#40/3#

Explanation:

Use the trig substitution #x = 4sintheta#. Then #dx = 4costhetad theta#. We adjust the bounds of integration accordingly.

#int_0^sin(2sqrt(3)) (4sintheta)^3/sqrt(16 - (4sintheta)^2) * 4costheta d theta#

#int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16 - 16sin^2theta) * 4costheta d theta#

#int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16(1 - sin^2theta)) * 4costheta d theta#

#int_0^sin(2sqrt(3)) (64sin^3theta)/sqrt(16cos^2theta) * 4costheta d theta#

#int_0^sin(2sqrt(3)) (64sin^3theta)/(4costheta) * 4costheta d theta#

#int_0^sin(2sqrt(3)) 64sin^3theta d theta#

#int_0^sin(2sqrt(3)) 64(1 - cos^2theta)sin theta d theta#

Now let #u = costheta#. Then #du = -sintheta d theta# and #d theta = (du)/(-sintheta)#.

#int_1^cos(sin(2sqrt(3))) 64(1 - u^2)sintheta * (du)/(-sintheta)#

#-int_1^cos(sin(2sqrt(3))) 64(1 - u^2)du#

#-int_1^cos(sin(2sqrt(3))) 64 - 64u^2 du#

#-[64u - 64/3u^3]_1^cos(sin(2sqrt(3))#

#-[64costheta - 64/3cos^3theta]_0^sin(2sqrt(3))#

#-[16sqrt(16 - x^2) - 1/3(16 - x^2)^(3/2)]_0^(2sqrt(3)#

#-(16sqrt(16 - (2sqrt(3))^2) - 1/3sqrt(16 - (2sqrt(3))^2)^(3/2) - (16sqrt(16 - 0) - 1/3(16 - 0)^(3/2)))#

#-16sqrt(4) + 1/3(4)^(3/2) + 16(4) - 1/3(64)#

#-32 + 8/3 + 64 - 64/3#

#40/3#

Hopefully this helps!


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