Find the Integral square root of 16-x^2

Problem

(∫_^)(√(,16−x^2)*d(x))

Solution

1. Identify the form of the integrand as √(,a^2−x^2) where a=4 which suggests using the trigonometric substitution x=4*sin(θ)

2. Differentiate the substitution to find d(x)=4*cos(θ)*d(θ)

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

(∫_^)(√(,16−(4*sin(θ))^2)⋅4*cos(θ)*d(θ))

4. Simplify the expression inside the square root using the identity 1−sin^2(θ)=cos^2(θ)

(∫_^)(√(,16*(1−sin^2(θ)))⋅4*cos(θ)*d(θ))

(∫_^)(4*cos(θ)⋅4*cos(θ)*d(θ))

(∫_^)(16*cos^2(θ)*d(θ))

5. Apply the power-reduction identity cos^2(θ)=(1+cos(2*θ))/2 to rewrite the integral.

(∫_^)(16⋅(1+cos(2*θ))/2*d(θ))

(∫_^)((8+8*cos(2*θ))*d(θ))

6. Integrate with respect to θ

8*θ+4*sin(2*θ)+C

7. Expand the double angle using sin(2*θ)=2*sin(θ)*cos(θ)

8*θ+8*sin(θ)*cos(θ)+C

8. Back-substitute to return to the variable x Since x=4*sin(θ) then sin(θ)=x/4 θ=arcsin(x/4) and cos(θ)=√(,1−(x/4)^2)=√(,16−x^2)/4

8*arcsin(x/4)+8*(x/4)*(√(,16−x^2)/4)+C

9. Simplify the final expression.

8*arcsin(x/4)+(x√(,16−x^2))/2+C

Final Answer

(∫_^)(√(,16−x^2)*d(x))=(x√(,16−x^2))/2+8*arcsin(x/4)+C

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