anonymous
  • anonymous
please helpppppppp... question in comment
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
wht do u neep help with
anonymous
  • anonymous
anonymous
  • anonymous
the exact value is

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More answers

IrishBoy123
  • IrishBoy123
re-write the integral using the identity and replacing the cube term
anonymous
  • anonymous
unable to do..
IrishBoy123
  • IrishBoy123
if A = B - C \[\int B = \int A+C\]
anonymous
  • anonymous
the intergal will replace the cube term
anonymous
  • anonymous
i need detailed explanation if you dont mind !!!!!!!!
anonymous
  • anonymous
im talking bout u need to take 1/2 and replace it with the cube term which is 3
anonymous
  • anonymous
you have to integrate first????
anonymous
  • anonymous
ik thts wht i was telling u
anonymous
  • anonymous
will you help in donig it ?
anonymous
  • anonymous
who me
anonymous
  • anonymous
@phi
anonymous
  • anonymous
@mbma526
phi
  • phi
do you know how to integrate cos 3x ?
anonymous
  • anonymous
no
phi
  • phi
do you know how to integrate cos x ?
anonymous
  • anonymous
no @phi
phi
  • phi
do you know the derivative of sin x ?
anonymous
  • anonymous
-cosx
phi
  • 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 \[ \int \cos x \ dx = \sin x + C\]
anonymous
  • anonymous
but for \[\cos ^{3} x\] ??
phi
  • 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
  • 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
  • phi
yes, there is a power of 3. patience.
phi
  • 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 ?
anonymous
  • anonymous
no not been able since 1 hour -___-
phi
  • phi
if you had 3= 4x + y can you solve for x ?
anonymous
  • anonymous
yess x=(3-y)/4
phi
  • phi
use that same idea to "solve" (which means isolate to one side) cos^3 x start with \[ \cos 3x =4 \cos^3 x - 3 \cos x \]
anonymous
  • anonymous
didnt read the question well.. thank you @phi
phi
  • phi
you should get \[ \cos^3 x = \frac{1}{4}\cos 3x + \frac{3}{4} \cos x \]
anonymous
  • anonymous
yesss thank you
phi
  • phi
now the integral \[ \int \cos^3 x \ dx \] can be written as \[ \frac{1}{4}\int \cos 3x \ dx +\frac{3}{4} \int \cos x \ dx\]
anonymous
  • anonymous
yess i can do it further.. thank you
phi
  • phi
yw

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