dfresenius
Integral with trig functions



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dfresenius
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dw:1361089150658:dw

dumbcow
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i am intrigued

dfresenius
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dfresenius
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or would it be cos^xcos5x

dumbcow
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no cant combine them...it would just be
\[\cos(5x)  \cos^{2} x \cos(5x)\]

dfresenius
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ok

dfresenius
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so left integral is easy

dfresenius
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hmm no i cant do that

dfresenius
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that 5x is just blowing my mind

dumbcow
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yeah im not sure of way to do it without expanding out the cos(5x) ..... which would be messy

hartnn
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huh ?
integral of cos 5x is just sin 5x/5 +c .

dfresenius
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dw:1361089595564:dw

dfresenius
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yA i know that integral, but its the cos^2 that is messin me up

hartnn
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ok, you need to used
\(\sin^2x = (1 \cos2x )/2\)

dfresenius
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okkk

hartnn
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then, for cos 2x cos 7x
there is a standard formula for , cos A cos B =...

hartnn
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*cos 2x cos 5x


dfresenius
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i got it then :)

hartnn
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good!

dfresenius
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dfresenius
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ok so here is another, im thinking with this one i can multiply by a 1/2 and use the identity sin2x

hartnn
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sin x cos x = (sin 2x) /2
then we have formula for
sin A cos B

hartnn
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yes! good thought :)

dfresenius
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nice!!

dfresenius
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sami21
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for the first question by parts can also be used .
\[\Large \frac{1}{5}\sin^2(x)\sin(5x)\frac{2}{5}\int\limits \sin(x)\sin(5x)\cos(x)dx\]

hartnn
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sin A cos B =... ?

dfresenius
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oops

.Sam.
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\[( \sin A ) ( \sin B )=\frac{1}{2} (\cos (A B )\cos (A +B ))\]

.Sam.
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But there's still a lot of work if you do this

dfresenius
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dfresenius
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hartnn
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why  *minus* in between ?

dfresenius
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integral of sin is cosx

hartnn
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both of  from that you have factored out, right ? to make 1/4 ?

dfresenius
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well, before i took integral

dfresenius
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dw:1361090806081:dw

dfresenius
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those will both be negative right?

hartnn
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integral of those will be negative, yes, right.

dfresenius
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ok thats all i got, thank you so much both of you for helping me out, big test this friday

hartnn
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best of luck (Y)