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How do you evaluate the integral of #int (1 + cos 4x)^(3/2) dx#?

Answer 1

Ezra Adams

#int(1+cos4x)^(3/2)dx=sqrt2/8(3sin2x+sin6x)+"c"#

#int(1+cos4x)^(3/2)dx=int(1+2cos^2 2x-1)^(3/2)dx=int2sqrt2cos^3 2xdx=sqrt2/2int(2cos^2x-1)^3dx#

After a very long expansion, we get it into the following form

#(sqrt2)/2int3cos2x+cos6xdx=sqrt2/2(1/2sin2x+1/6sin6x)+"c"# #=sqrt2/8(3sin2x+sin6x)+"c"#

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Answer 2

Christopher Allers

To evaluate the integral (\int (1 + \cos(4x))^{3/2} , dx), you can use the trigonometric identity (\cos^2(x) = \frac{1 + \cos(2x)}{2}) to simplify the integral. Let (u = 1 + \cos(4x)), then (du = -4\sin(4x) , dx). Using the identity, you can rewrite (\cos(4x)) as (1 - 2\sin^2(2x)). Substituting (u) and (du) into the integral and simplifying will yield a form suitable for integration. You'll then integrate with respect to (u) and back-substitute to find the final answer.

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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.