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SOLUTION 16: Integrate ∫√1−x⋅√x+3 dx. Begin by rewriting the integrand.
∫√1−x⋅√x+3 dx=∫√−x2−2x+3 dx
=∫√−(x2+2x)+3 dx
(Now complete the square.)
=∫√−(x2+2x+1)+3+1 dx
=∫√4−(x+1)2 dx
Now use the trig substitution
x+1=2sinθ
so that
dx=2cosθ dθ
Substitute into the original problem, replacing all
forms of x, getting
∫√4−(x+1)2 dx=∫√4−(2sinθ)2⋅2cosθ dθ
=2∫√4−4sin2θcosθ dθ
=2∫√4(1−sin2θ)cosθ dθ
=2∫√4cos2θcosθ dθ
=2∫2cosθ⋅cosθ dθ
=4∫cos2θ dθ
(Recall that cos2θ=(1/2)(1+cos2θ).)
=4∫12(1+cos2θ) dθ
=2∫(1+cos2θ) dθ
=2(θ+12sin2θ)+C
(Recall that sin2θ=2sinθcosθ.)
=2(θ+12(2sinθcosθ))+C
=2(θ+sinθcosθ)+C
=2θ+2sinθcosθ+C
( We need to write our final answer in terms of x.
Since
x+1=2sinθ
it follows that
sinθ=x+12=oppositehypotenuse
and
θ=arcsinx+12
and from the Pythagorean Theorem that
(adjacent)2+(opposite)2=(hypotenuse)2 ⟶
(adjacent)2+(x+1)2=(2)2 ⟶ adjacent=√4−(x+1)2=√−x2−2x+3 ⟶
cosθ=adjacenthypotenuse=√−x2−2x+32.)
=2arcsinx+12+2⋅x+12√−x2−2x+32
=2arcsinx+12+x+12√−x2−2x+3
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