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What is the integral of #int (sin2x)/sinx #?

Answer 1

William Abdalla

#= 2 sinx + C#

#int dx qquad (sin2x)/sinx#

#int dx qquad (2 sinx cos x)/sinx#

#2 int dx qquad cos x#

#= 2 sinx + C#

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Answer 2

Elijah Abram

The integral of (\int \frac{\sin(2x)}{\sin(x)} , dx) is (-2\ln|\csc(x) + \cot(x)| + C), where (C) is the constant of integration.

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Answer 3

Ezra Adams

To find the integral of (\frac{\sin(2x)}{\sin(x)}), we can use the following steps:

Rewrite the numerator using the double angle identity: (\sin(2x) = 2\sin(x)\cos(x)).
Rewrite the integral as: (\int \frac{2\sin(x)\cos(x)}{\sin(x)} , dx).
Cancel out the (\sin(x)) term in the denominator.
Integrate (\cos(x)) with respect to (x).

Following these steps, the integral becomes:

[\int 2\cos(x) , dx]

The integral of (\cos(x)) is (\sin(x)) (plus a constant), so integrating (2\cos(x)) with respect to (x) gives:

[2\sin(x) + C]

Where (C) is the constant of integration. Therefore, the integral of (\frac{\sin(2x)}{\sin(x)}) with respect to (x) is (2\sin(x) + 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.