Antiderivative of 1 4 </mfrac> sin &#x2061;<!-- ⁡ --> ( 2 x )

Summer Bradford 2022-06-25 Answered
Antiderivative of 1 4 sin ( 2 x )
I am trying to find the antiderivative of 1 4 sin ( 2 x ) but i'mn ot sure how to find this. Can someone maybe give me what formula to use when it involves sin?
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Answers (2)

Blaze Frank
Answered 2022-06-26 Author has 18 answers
Step 1
The beauty of trig is that you can get wildly different looking results depending on how you start. For example take identify: sin ( 2 x ) = 2 sin ( x ) cos ( x )
Then you can write 1 4 sin ( 2 x ) d x = 1 2 sin ( x ) cos ( x ) d x .
Set u = cos ( x ) then d u = sin ( x ) d x. So now we solve the indefinite integral: 1 2 u d u = 1 4 u 2 + C
Substituting back in for u = cos ( x ) we get
1 4 sin ( 2 x ) d x = 1 4 cos 2 ( x ) + C
This looks very different but you can see that it's equivalent to the first answer given to your question by remembering the following:
cos 2 ( x ) = 1 sin 2 ( x )
and sin 2 ( x ) = 1 cos ( 2 x ) 2 and finally keeping in mind that C of the two answers is not the same.

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Arraryeldergox2
Answered 2022-06-27 Author has 10 answers
Step 1
I := 1 4 sin ( 2 x )   d x
I = 1 4 sin ( 2 x )   d x
Let u = 2 x , d u = 2   d x d u 2 = d x.
I = 1 4 1 2 sin ( u )   d u
I = 1 8 cos u + C
I = 1 8 cos ( 2 x ) + C
Why is this correct? We can take the derivative of our answer to see we get our original function back and that this and our primitive are equal up to a constant.
d d x ( 1 8 cos ( 2 x ) ) = 1 8 sin ( 2 x ) 2 = 1 4 sin ( 2 x )
It is helpful to note these few facts to answer your question about what to do when you have sin or cos in an integrand:
sin x   d x = cos x + C
cos x   d x = sin x + C
Note that this just involves x, but you can still do your normal u-substitutions as necessary to get it in the form above, as I did.

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