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
$a^2=9anda=3$

$b^2=1andb=1$

$x=a/b*sintheta=sintheta$

$dx=costheta*d(theta)$

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

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

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

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

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

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

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

$theta=sin^-1x$

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