Evaluate the Integral integral of sin(x)^6 with respect to x

Problem

(∫_^)(sin6(x)*d(x))

Solution

1. Apply power-reduction formula for sin2(x) to rewrite the integrand.

sin2(x)=(1−cos(2*x))/2

2. Rewrite the expression as the cube of the squared term.

(∫_^)(sin6(x)*d(x))=(∫_^)((sin2(x))3*d(x))

3. Substitute the identity into the integral.

(∫_^)(((1−cos(2*x))/2)3*d(x))=1/8*(∫_^)((1−cos(2*x))3*d(x))

4. Expand the binomial (1−cos(2*x))3

1/8*(∫_^)((1−3*cos(2*x)+3*cos2(2*x)−cos3(2*x))*d(x))

5. Apply power-reduction to cos2(2*x) and use the identity cos3(2*x)=cos(2*x)*(1−sin2(2*x))

cos2(2*x)=(1+cos(4*x))/2

cos3(2*x)=cos(2*x)−cos(2*x)*sin2(2*x)

6. Substitute these identities back into the integral.

1/8*(∫_^)((1−3*cos(2*x)+3/2*(1+cos(4*x))−(cos(2*x)−cos(2*x)*sin2(2*x)))*d(x))

7. Simplify the integrand by combining like terms.

1/8*(∫_^)((5/2−4*cos(2*x)+3/2*cos(4*x)+cos(2*x)*sin2(2*x))*d(x))

8. Integrate term by term, using the substitution u=sin(2*x) for the last term.

1/8*(5/2*x−2*sin(2*x)+3/8*sin(4*x)+1/6*sin3(2*x))+C

9. Distribute the constant to find the final expression.

5/16*x−1/4*sin(2*x)+3/64*sin(4*x)+1/48*sin3(2*x)+C

Final Answer

(∫_^)(sin6(x)*d(x))=(5*x)/16−sin(2*x)/4+(3*sin(4*x))/64+sin3(2*x)/48+C

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