Final answer:
To evaluate the given integrals using u-substitution, we can substitute u = sin(2x) for the first two integrals and u = sin(x) for the last integral. After integrating with respect to u and substituting back the original variable, we obtain the answers (1/8)sin¹&sup8;(2x) + C, (1/4)sin´(2x) + C, and (1/10)sin¹&sup8;(x) + C.
Step-by-step explanation:
To evaluate the integral ∫ sin¹&sup7;(2x) cos(2x) dx, we can use the substitution u = sin(2x). This means that du = 2cos(2x) dx. So the integral becomes ∫ u¹&sup7; du = (1/8)u¹&sup8; + C, where C is the constant of integration. Substituting back u = sin(2x), we get (1/8)sin¹&sup8;(2x) + C.
Similarly, to evaluate the integral ∫ sin³(2x) dx, we can use the substitution u = sin(2x). This means that du = 2cos(2x) dx. So the integral becomes ∫ u³ du = (1/4)u´ + C. Substituting back u = sin(2x), we get (1/4)sin´(2x) + C.
Finally, to evaluate the integral ∫ sin¹&sup8;(x) dx, we can use the substitution u = sin(x). This means that du = cos(x) dx. So the integral becomes ∫ u¹&sup7; du = (1/10)u¹&sup8; + C. Substituting back u = sin(x), we get (1/10)sin¹&sup8;(x) + C.