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Integral with trig functions

Mathematics
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|dw:1361089150658:dw|
i am intrigued
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Other answers:

or would it be cos^xcos5x
no cant combine them...it would just be \[\cos(5x) - \cos^{2} x \cos(5x)\]
ok
so left integral is easy
hmm no i cant do that
that 5x is just blowing my mind
yeah im not sure of way to do it without expanding out the cos(5x) ..... which would be messy
huh ? integral of cos 5x is just sin 5x/5 +c .
|dw:1361089595564:dw|
yA i know that integral, but its the cos^2 that is messin me up
ok, you need to used \(\sin^2x = (1- \cos2x )/2\)
okkk
then, for cos 2x cos 7x there is a standard formula for , cos A cos B =...
*cos 2x cos 5x
ahh i always forget sum to product rules ... anyway, here is solution http://www.wolframalpha.com/input/?i=integrate+sin^2%28x%29+cos%285x%29+dx
i got it then :)
good!
|dw:1361090139384:dw|
ok so here is another, im thinking with this one i can multiply by a 1/2 and use the identity sin2x
sin x cos x = (sin 2x) /2 then we have formula for sin A cos B
yes! good thought :)
nice!!
|dw:1361090244648:dw|
for the first question by parts can also be used . \[\Large \frac{1}{5}\sin^2(x)\sin(5x)-\frac{2}{5}\int\limits \sin(x)\sin(5x)\cos(x)dx\]
sin A cos B =... ?
oops
\[( \sin A ) ( \sin B )=\frac{1}{2} (\cos (A -B )-\cos (A +B ))\]
But there's still a lot of work if you do this
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|dw:1361090501054:dw|
why - *minus* in between ?
integral of sin is -cosx
both of - from that you have factored out, right ? to make -1/4 ?
well, before i took integral
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those will both be negative right?
integral of those will be negative, yes, right.
ok thats all i got, thank you so much both of you for helping me out, big test this friday
best of luck (Y)

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