please helpppppppp... question in comment

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please helpppppppp... question in comment

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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wht do u neep help with
the exact value is

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Other answers:

re-write the integral using the identity and replacing the cube term
unable to do..
if A = B - C \[\int B = \int A+C\]
the intergal will replace the cube term
i need detailed explanation if you dont mind !!!!!!!!
im talking bout u need to take 1/2 and replace it with the cube term which is 3
you have to integrate first????
ik thts wht i was telling u
will you help in donig it ?
who me
  • phi
do you know how to integrate cos 3x ?
no
  • phi
do you know how to integrate cos x ?
no @phi
  • phi
do you know the derivative of sin x ?
-cosx
  • phi
just 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 \[ \int \cos x \ dx = \sin x + C\]
but for \[\cos ^{3} x\] ??
  • phi
let'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
that 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
yes, there is a power of 3. patience.
  • phi
hopefully 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 ?
no not been able since 1 hour -___-
  • phi
if you had 3= 4x + y can you solve for x ?
yess x=(3-y)/4
  • phi
use that same idea to "solve" (which means isolate to one side) cos^3 x start with \[ \cos 3x =4 \cos^3 x - 3 \cos x \]
didnt read the question well.. thank you @phi
  • phi
you should get \[ \cos^3 x = \frac{1}{4}\cos 3x + \frac{3}{4} \cos x \]
yesss thank you
  • phi
now the integral \[ \int \cos^3 x \ dx \] can be written as \[ \frac{1}{4}\int \cos 3x \ dx +\frac{3}{4} \int \cos x \ dx\]
yess i can do it further.. thank you
  • phi
yw

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