Final answer:
The solution to the integral is: (1/2)(2u + (1/4)Sin(4x))⋅[2+Sin(2x)]⁵ - (1/4)Cos(4x)⋅[2+Sin(2x)]⁶ - (1/6)⋅Cos(4x)⋅[2+Sin(2x)]⁶ + C.
Step-by-step explanation:
To solve the given integral ∫Cos³(2x)⋅[2+Sin(2x)]⁵dx, we can use the substitution method.
Here's how:
1. Let's make a substitution: u = Sin(2x). Taking the derivative of both sides with respect to x, we get du/dx = 2Cos(2x).
2. Rearranging the equation, we have dx = (1/2)du/Cos(2x).
3. Now, substitute u = Sin(2x) and dx = (1/2)du/Cos(2x) into the integral.
4. The integral becomes ∫(Cos³(2x)⋅[2+Sin(2x)]⁵)(1/2)du/Cos(2x).
5. Simplify by canceling out the Cos(2x) terms in the numerator and denominator: ∫(Cos²(2x)⋅[2+Sin(2x)]⁵)(1/2)du.
6. Next, simplify further: ∫(1/2)Cos²(2x)⋅[2+Sin(2x)]⁵du.
7. Expand the Cos²(2x) term using the identity Cos²(2x) = (1/2)(1 + Cos(4x)): ∫(1/2)(1 + Cos(4x))⋅[2+Sin(2x)]⁵du.
8. Distribute and simplify: ∫(1/2)(2 + Cos(4x) + 2⋅Cos(4x)⋅Sin(2x) + Cos(4x)⋅Sin²(2x))⋅[2+Sin(2x)]⁵du.
9. Now, integrate each term separately with respect to u:
- - ∫(1/2)(2 + Cos(4x))⋅[2+Sin(2x)]⁵du = (1/2)(2u + (1/4)Sin(4x))⋅[2+Sin(2x)]⁵ + C₁
- - ∫(1/2)(2⋅Cos(4x)⋅Sin(2x))⋅[2+Sin(2x)]⁵du = -(1/4)Cos(4x)⋅[2+Sin(2x)]⁶ + C₂
- - ∫(1/2)(Cos(4x)⋅Sin²(2x))⋅[2+Sin(2x)]⁵du = -(1/6)⋅Cos(4x)⋅[2+Sin(2x)]⁶ + C₃
10. Finally, combining all the terms, the solution to the integral is:
(1/2)(2u + (1/4)Sin(4x))⋅[2+Sin(2x)]⁵ - (1/4)Cos(4x)⋅[2+Sin(2x)]⁶ - (1/6)⋅Cos(4x)⋅[2+Sin(2x)]⁶ + C,
where C = C₁ + C₂ + C₃ is the constant of integration.
Your question is incomplete, but most probably the full question was:
Solve the integral and explain .
I(12π)= ∫Cos³(2x)⋅[2+Sin(2x)]⁵dx