How to solve ∫x2√9−x2dx with integration by parts?

Does it involve trig substitution?

1 Answer
James
May 2, 2018

the answer ∫sin2θ−sin4θ⋅d(θ)=−sin(4sin−1(x))−4⋅sin−1(x)32

Explanation:

show the steps

∫x2√9−x2dx

suppose
a2=9anda=3

b2=1andb=1

x=ab⋅sinθ=sinθ

dx=cosθ⋅d(θ)

∫sin2θ⋅cosθ⋅cosθ⋅d(θ)

∫sin2θ⋅cos2θ⋅d(θ)

∫sin2θ⋅(1−sin2θ)⋅d(θ)

∫sin2θ−sin4θ⋅d(θ)

∫sin2θ⋅d(θ)=12[θ−cosθ⋅sinθ]

∫sin4θ⋅d(θ)=sin(4⋅θ)−8⋅sin(2⋅θ)+12⋅θ32

∫sin2θ−sin4θ⋅d(θ)=−sin(4θ)−4⋅θ32

θ=sin−1x

∫sin2θ−sin4θ⋅d(θ)=−sin(4sin−1(x))−4⋅sin−1(x)32

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