How to solve #\intx^2\sqrt(9-x^2)dx# with integration by parts?

Question

Does it involve trig substitution?

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Answer 1 · 2018-05-02T00:53:09

the answer #intsin^2theta-sin^4theta*d(theta)=-(sin(4sin^-1(x))-4*sin^-1(x))/32#

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#intx^2\sqrt(9-x^2)dx#

suppose
#a^2=9anda=3#

#b^2=1andb=1#

#x=a/b*sintheta=sintheta#

#dx=costheta*d(theta)#

#intsin^2theta*costheta*costheta*d(theta)#

#intsin^2theta*cos^2theta*d(theta)#

#intsin^2theta*(1-sin^2theta)*d(theta)#

#intsin^2theta-sin^4theta*d(theta)#

#intsin^2theta*d(theta)=1/2[theta-costheta*sintheta]#

#intsin^4theta*d(theta)=(sin(4*theta)-8*sin(2*theta)+12*theta)/32#

#intsin^2theta-sin^4theta*d(theta)=-(sin(4theta)-4*theta)/32#

#theta=sin^-1x#

#intsin^2theta-sin^4theta*d(theta)=-(sin(4sin^-1(x))-4*sin^-1(x))/32#