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How do you evaluate #intsin(2x)/sin(x)dx#?

Answer 1

Luke Alvarez

Use #sin(2x)=2sin(x)cos(x)#.

What you need is a double angle trig. identity. Namely, #sin(2x)=2sin(x)cos(x)# will do the trick.

#intsin(2x)/sin(x)dx=>int(2sin(x)cos(x))/sin(x)dx#

Now you can get #sin(x)# to cancel, leaving you with:

#int2cos(x)dx=2sin(x)+C#

Good luck!

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Answer 2

Charles Anctil

To evaluate the integral ∫(sin(2x)/sin(x)) dx, you can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the integrand as 2cos(x). Thus, the integral becomes ∫(2cos(x)) dx. Integrating 2cos(x) with respect to x gives 2sin(x) + C, where C is the constant of integration. Therefore, the integral of sin(2x)/sin(x) with respect to x is 2sin(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.