Evaluate the Integral

Problem

(∫_^)((x2)/√(,16−x2)*d(x))

Solution

1. Identify the form of the integrand, which contains √(,a2−x2) where a=4

2. Apply the substitution x=4*sin(θ) which implies d(x)=4*cos(θ)*d(θ)

3. Substitute these into the integral to transform it into a trigonometric integral.

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

4. Simplify the denominator using the identity 1−sin2(θ)=cos2(θ)

√(,16−16*sin2(θ))=√(,16*cos2(θ))=4*cos(θ)

5. Reduce the integral by canceling the 4*cos(θ) terms.

(∫_^)((16*sin2(θ))/(4*cos(θ))⋅4*cos(θ)*d(θ))=(∫_^)(16*sin2(θ)*d(θ))

6. Apply the power-reduction identity sin2(θ)=(1−cos(2*θ))/2

(∫_^)(16⋅(1−cos(2*θ))/2*d(θ))=(∫_^)((8−8*cos(2*θ))*d(θ))

7. Integrate with respect to θ

8*θ−4*sin(2*θ)+C

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

8*θ−8*sin(θ)*cos(θ)+C

9. 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)=√(,16−x2)/4

8*arcsin(x/4)−8*(x/4)*(√(,16−x2)/4)+C

10. Simplify the final expression.

8*arcsin(x/4)−(x√(,16−x2))/2+C

Final Answer

(∫_^)((x2)/√(,16−x2)*d(x))=8*arcsin(x/4)−(x√(,16−x2))/2+C

Want more problems? Check here!