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How do you find the integral of #int [cos^3 (2x)] dx #?

Answer 1

Ezra Adams

#1/2sin 2x-1/6sin^3 2x+C#

#I=int cos^3 2xdx = int cos 2x cos^2 2x dx# #I=int cos 2x (1-sin^2 2x)dx#

#sin 2x=t => 2cos 2xdx = dt => cos 2xdx = dt/2#

#I=int(1-t^2)dt/2=1/2(t-t^3/3)+C#

#I=1/2sin 2x-1/6sin^3 2x+C#

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Answer 2

Christopher Allers

To find the integral of ∫cos^3(2x) dx, you can use the trigonometric identity cos^3(x) = (1/4)(3cos(x) + cos(3x)). So, the integral becomes ∫(1/4)(3cos(2x) + cos(6x)) dx. You can then integrate each term separately. The integral of 3cos(2x) dx is (3/2)sin(2x), and the integral of cos(6x) dx is (1/6)sin(6x). Therefore, the final result is (3/2)sin(2x) + (1/6)sin(6x) + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.