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

Does it involve trig substitution?

Answer 1

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

show the steps

#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#
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Answer 2

To solve the integral ∫x^2√(9-x^2) dx using integration by parts, follow these steps:

  1. Let u = x^2 and dv = √(9 - x^2) dx.
  2. Compute du and v.
  3. Apply the integration by parts formula: ∫u dv = uv - ∫v du.
  4. Substitute the values of u, dv, du, and v into the integration by parts formula.
  5. Evaluate the resulting integral.
  6. Simplify the expression to obtain the final answer.
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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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