Evaluate the Integral

Problem

(∫_^)(√(,36−x2)*d(x))

Solution

1. Identify the form of the integrand as √(,a2−x2) where a=6 which suggests using the trigonometric substitution x=6*sin(θ)

2. Differentiate the substitution to find d(x)=6*cos(θ)*d(θ)

3. Substitute the expressions for x and d(x) into the integral:

(∫_^)(√(,36−(6*sin(θ))2)⋅6*cos(θ)*d(θ))

4. Simplify the term inside the square root using the identity 1−sin2(θ)=cos2(θ)

(∫_^)(√(,36*(1−sin2(θ)))⋅6*cos(θ)*d(θ))

(∫_^)(6*cos(θ)⋅6*cos(θ)*d(θ))

(∫_^)(36*cos2(θ)*d(θ))

5. Apply the power-reduction identity cos2(θ)=(1+cos(2*θ))/2

(∫_^)(36⋅(1+cos(2*θ))/2*d(θ))

18*(∫_^)((1+cos(2*θ))*d(θ))

6. Integrate with respect to θ

18*(θ+sin(2*θ)/2)+C

18*θ+9*sin(2*θ)+C

7. Use the double-angle identity sin(2*θ)=2*sin(θ)*cos(θ) to prepare for back-substitution:

18*θ+18*sin(θ)*cos(θ)+C

8. Back-substitute using sin(θ)=x/6 θ=arcsin(x/6) and cos(θ)=√(,36−x2)/6

18*arcsin(x/6)+18*(x/6)*(√(,36−x2)/6)+C

9. Simplify the final expression:

18*arcsin(x/6)+(x√(,36−x2))/2+C

Final Answer

(∫_^)(√(,36−x2)*d(x))=18*arcsin(x/6)+(x√(,36−x2))/2+C

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