+0

# is this answer wrong !

0
213
6
+1828

I think it must be sin4(x-3) insted of cos4(x-3) in the last step !

xvxvxv  Feb 13, 2015

#5
+26399
+10

Yes, you could do, as there is a "c" to wrap up the constants (though you would change the cos term to a sin term as per your original question!)

.

Alan  Feb 13, 2015
Sort:

#1
+18827
+10

I think it must be sin4(x-3) insted of cos4(x-3) in the last step !

Yes!

Because: $$\int{\cos(4u)}\ du = \frac{1}{4}\sin(4u)$$

heureka  Feb 13, 2015
#2
+1828
0

Also in this question, why they didn't wrote $$\frac{1}{2}(x+1)$$  insted of $$\frac{1}{2}x$$ !

xvxvxv  Feb 13, 2015
#3
+26399
+10

$$\frac{1}{2}\int (1+\cos{2(x+1)})dx=\frac{1}{2}\int dx+\frac{1}{2}\int \cos{2(x+1)}dx$$

so the single x comes from the first integral on the right-hand side. However, you could add whatever constant you like, because, as it's an indefinite integral, any other constants are accounted for in the "c" at the end.

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Alan  Feb 13, 2015
#4
+1828
0

So here can I wrote $$\frac{3}{8}x$$ insted of $$\frac{3}{8}(x-3)$$  ?

xvxvxv  Feb 13, 2015
#5
+26399
+10

Yes, you could do, as there is a "c" to wrap up the constants (though you would change the cos term to a sin term as per your original question!)

.

Alan  Feb 13, 2015
#6
+1828
+5

very clear

thank you Alan and heureka

xvxvxv  Feb 13, 2015

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