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Mathematics 20 Online
OpenStudy (anonymous):

Find intergral sin(x)cos(x)

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.

OpenStudy (baldymcgee6):

http://screencast.com/t/g6dNJGxjezEf

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.

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)**

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.

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.

Parth (parthkohli):

And you're welcome. :)

Parth (parthkohli):

<Vulcan greet>

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