Integrate ? Sin 2x dx

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Answer 1 · 2018-04-06T05:15:21

#intsin2xdx=-1/2cos2x+C#

We want #intsin2xdx#.

In general, #intsinaxdx=-1/acosax+C#. A proof of this will be shown.

So,

#intsin2xdx=-1/2cos2x+C#

Proof:

Let #u=2x#.

#(du)/dx=2#

#du=2dx#

Solve for #dx:#

#1/2du=dx#

Apply the substitution, factoring the constant #1/2# outside of the integral.

#1/2intsinudu=-1/2cosu+C#

Recalling that #u=2x:#

#intsin2xdx=-1/2cos2x+C#

Answer 2 · 2018-04-06T08:20:14

alternate...

# int sin2x dx = 2 int sinx cosx dx #

#u = sinx #

#=> du = cosx dx #

#=> 2 int u du #

#=> u^2 + c #

#=> color(blue)(sin^2 x + c #