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What is the integral of #int sin(2x)dx#?

Answer 1

Emilia Aguilar

#-1/2cos(2x)+C#

We will use the rule:

#intsin(u)du=-cos(u)+C#

So, if we have

#intsin(2x)dx#

we should set

#u=2x" "=>" "du=2dx#

In order to have a #2dx# in the integral expression, multiply by a #2# on the inside and balance it with a #1/2# on the outside:

#=1/2intsin(2x)*2dx#

Substitute:

#=1/2intsin(u)du#

Using the initial rule, this becomes:

#=-1/2cos(u)+C#

Since #u=2x#,

#=-1/2cos(2x)+C#

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Answer 2

Adam Adkins

The integral of sin(2x) with respect to x is (-1/2)cos(2x) + C, where C is the constant of integration.

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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.