How do you integrate #sqrt(1+(4cos^2))dx#?

Answer 1
Normally, you might expect that the #1# be a #4# so you can just do trig substitution. Unfortunately we can't do that.

If you are not in a very advanced Calculus course, where you might work with elliptic integrals, you don't have to know how to do this analytically. At best you can do a numeric integral.

So, you can evaluate #int_a^b sqrt(1+4cos^2x)dx# on your calculator or on Wolfram Alpha, but it's not something I believe you have been taught to do by hand.
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Answer 2

To integrate √(1 + 4cos^2(x)) dx, you can use the trigonometric identity cos^2(x) = (1 + cos(2x))/2:

√(1 + 4cos^2(x)) dx = √(1 + 4(1 + cos(2x))/2) dx = √(1 + 2 + 2cos(2x)) dx = √(3 + 2cos(2x)) dx

Now, let u = 2x, then du = 2dx:

= (1/2)√(3 + cos(u)) du

Now, integrate (1/2)√(3 + cos(u)) du with respect to u, then substitute back u = 2x to obtain the final result.

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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.

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.

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.

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.

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