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How do you integrate #sin(3x)cos(3x)dx#?

Answer 1

Emma Aldridge

Try this:

hope it helps

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Answer 2

Ezekiel Alexander

To integrate sin(3x)cos(3x)dx, you can use the trigonometric identity:

sin(a)cos(b) = 1/2 * [sin(a + b) + sin(a - b)]

Applying this identity to the given expression, we get:

sin(3x)cos(3x) = 1/2 * [sin(3x + 3x) + sin(3x - 3x)] = 1/2 * [sin(6x) + sin(0)] = 1/2 * [sin(6x) + 0] = 1/2 * sin(6x)

So, the integral becomes:

∫ sin(3x)cos(3x)dx = ∫ 1/2 * sin(6x) dx = 1/2 * ∫ sin(6x) dx

Now, integrate sin(6x) with respect to x:

∫ sin(6x) dx = -1/6 * cos(6x) + C

Where C is the constant of integration.

Therefore,

∫ sin(3x)cos(3x)dx = 1/2 * (-1/6 * cos(6x)) + C = -1/12 * cos(6x) + C

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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.