Mathematics
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OpenStudy (anonymous):
Find intergral sin(x)cos(x)
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hartnn (hartnn):
write sin x cos x as sin 2x / 2
and integrate
Parth (parthkohli):
First step is to take the constant out which is?
OpenStudy (anonymous):
why sin2x /2? is that one of the trig identity
hartnn (hartnn):
sin2x = 2 sin x cos x is a standard identity
Parth (parthkohli):
Yes.
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Parth (parthkohli):
\[\sin (2x) = 2\sin x \cos x\]Divide both sides by 2.
OpenStudy (anonymous):
but there is a 2..u can just divide it out..?
OpenStudy (anonymous):
ohh..dont know i can do that.. ok thanks everyone
Parth (parthkohli):
No, but you can take it outside.
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Parth (parthkohli):
sin(2x)/2 = 1/2 * sin(2x)
Since 1/2 is a constant, you can take it out of the integral.
OpenStudy (anonymous):
oh ok thanks. i get it
Parth (parthkohli):
Now, what's the integral of sin(2x)?
OpenStudy (anonymous):
-cosx
Parth (parthkohli):
-cos(2x)**
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OpenStudy (anonymous):
oh , 2x, then just 1/2
Parth (parthkohli):
-cos(2x) + C**
hartnn (hartnn):
or u can do what @baldymcgee6 has posted
Parth (parthkohli):
\[\rm \left({1 \over 2} \times -\cos(2x) \right)+ C\]
Parth (parthkohli):
@hartnn He hasn't done it. It's WolframAlpha.
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Parth (parthkohli):
It's a shame how people who use machines get medals.
OpenStudy (anonymous):
but cos(2x) is different from cos^2 (X) right
Parth (parthkohli):
Yes.
OpenStudy (anonymous):
haha, u guys all deserve medals :D really helpful
but wolframalpha 's answer is in cos^2(x)
here is in 2x
Parth (parthkohli):
Never mind wolframalpha.
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Parth (parthkohli):
And you're welcome. :)
Parth (parthkohli):
<Vulcan greet>