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what am I doing wrong in this trig integration question? (1 Viewer)

SureBluff

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for the sake of convenience, I'll call the integration sign :) ... (you know, the sign that looks like a giant 'f' or distorted 's')

find primitive of (sinx)^2

:)(sinx)^2dx
=:)1/2(1-cos2x)dx
=1/2:)(1-cosu)(1/2du)
=1/4(x-sin2x)+C

the answer at the back of the book was (1/2)x-(1/4)sin2x+C
I dunno where I went wrong, help!!!
 

shafqat

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SureBluff said:
for the sake of convenience, I'll call the integration sign :) ... (you know, the sign that looks like a giant 'f' or distorted 's')

find primitive of (sinx)^2

:)(sinx)^2dx
=:)1/2(1-cos2x)dx
=1/2:)(1-cosu)(1/2du)
= 1/4(u - cosu)
= 1/4(2x - sin2x)
= x/2 - 1/4 sin2x + C
the answer at the back of the book was (1/2)x-(1/4)sin2x+C
I dunno where I went wrong, help!!!
.......................
 
Last edited:

renya

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Dec 1, 2004
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SureBluff said:
=1/2:)(1-cosu)(1/2du)
your right up to this line
then u get
=1/4 :) (1 - cos u) du
=1/4 [u - sin u ] + C
=1/4 [2x - sin 2x] + C
=(1/2)x - (1/4)sin 2x + C
 

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