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anonymous
 one year ago
please helpppppppp...
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anonymous
 one year ago
please helpppppppp... question in comment

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wht do u neep help with

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0rewrite the integral using the identity and replacing the cube term

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0if A = B  C \[\int B = \int A+C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the intergal will replace the cube term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i need detailed explanation if you dont mind !!!!!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im talking bout u need to take 1/2 and replace it with the cube term which is 3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have to integrate first????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ik thts wht i was telling u

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0will you help in donig it ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to integrate cos 3x ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to integrate cos x ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2do you know the derivative of sin x ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2just cos x \[ \frac{d}{dx} \sin x = \cos x \] we can "undo the derivative" by integrating \[ \int d \sin x = \int \cos x \ dx \\ \sin x = \int \cos x \ dx \] I'm not sure that is clear, but people memorize this integral \[ \int \cos x \ dx = \sin x + C\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but for \[\cos ^{3} x\] ??

phi
 one year ago
Best ResponseYou've already chosen the best response.2let's first do cos 3x if we let u= 3x and we "work backwards" \[ \frac{d}{dx} \sin u = \cos u \frac{d}{dx} u = \cos 3x \cdot 3\\ \frac{d}{dx} \sin 3x = 3 \cos 3x \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2that is the derivative of sin 3x using the chain rule if we want to integrate cos 3x we need a 3 out front: 3 cos 3x and to compensate , we multiply by 1/3 \[ \frac{1}{3} \int 3 \cos 3x \ dx = \frac{1}{3} \sin 3x \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2yes, there is a power of 3. patience.

phi
 one year ago
Best ResponseYou've already chosen the best response.2hopefully you can integrate cos x and cos 3x (see above) we are given \[ \cos 3x =4 \cos^3 x  3 \cos x \] can you solve for cos^3 x ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no not been able since 1 hour ___

phi
 one year ago
Best ResponseYou've already chosen the best response.2if you had 3= 4x + y can you solve for x ?

phi
 one year ago
Best ResponseYou've already chosen the best response.2use that same idea to "solve" (which means isolate to one side) cos^3 x start with \[ \cos 3x =4 \cos^3 x  3 \cos x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0didnt read the question well.. thank you @phi

phi
 one year ago
Best ResponseYou've already chosen the best response.2you should get \[ \cos^3 x = \frac{1}{4}\cos 3x + \frac{3}{4} \cos x \]

phi
 one year ago
Best ResponseYou've already chosen the best response.2now the integral \[ \int \cos^3 x \ dx \] can be written as \[ \frac{1}{4}\int \cos 3x \ dx +\frac{3}{4} \int \cos x \ dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yess i can do it further.. thank you
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