How do you find the integral of #(x^2)/(16-x^2)^(1/2)#?

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Apr 20, 2018

Answer:

#intx^2/(16-x^2)^(1/2)dx=8arcsin(x/4)-(xsqrt(16-x^2))/2+C#

Explanation:

Rewrite with a root:

#intx^2/sqrt(16-x^2)dx#

For integrals involving the root #sqrt(a^2-x^2),# we use the substitution #x=asintheta.#

Here, #a^2=16, a=4, x=4sintheta, dx=4costhetad theta# and we get

#int(16sin^2theta4costhetad theta)/sqrt(16(1-sin^2theta))#

Recalling that #1-sin^2theta=cos^2theta,#we get

#=16int(sin^2thetacosthetad theta)/sqrt(cos^2theta)#

#=16int(sin^2thetacancelcostheta)/(cancelcostheta)d theta#

#=16intsin^2thetad theta#

Recalling that #sin^2theta=1/2(1-cos2theta),# we get

#16/2int(1-cos2thetad theta)=8(theta-1/2sin2theta)+C#

We need to get things in terms of #x#.

Recalling that #x=4sintheta, sintheta=x/4, theta=arcsin(x/4)#

To determine #1/2sin2theta,# recall the identity #1/2sin2theta=sinthetacostheta.# To use this, we need to determine the cosine using the below Pythagorean identity:

#sin^2theta+cos^2theta=1#
#cos^2theta=1-sin^2theta#
#cos^2theta=1-x^2/16#

#costheta=sqrt(16-x^2)/4#

Thus, #1/2sin2theta=(x/4)sqrt(16-x^2)/4=(xsqrt(16-x^2))/16#

And

#intx^2/(16-x^2)^(1/2)dx=8arcsin(x/4)-(xsqrt(16-x^2))/2+C#

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