Integrate cos^3xdx - First Clue: Substitute (1-(Sinx)^2)cosxdx

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In summary, the conversation discusses methods for integrating \int cos^3xdx and \int (1 - \sin^2 x) \cos x \, dx. The first suggestion is to substitute (1-sin^2 x)cos xdx, and the conversation continues to discuss different techniques that may be applicable. The conversation also includes a request for someone to check the validity of the work for \int cos^5[x]dx, which involves substituting (1-sin^2[x])^2 for cos^4[x] and using the substitution u=sinx to simplify the integral. The conversation concludes with a comment about using a general reduction formula.
  • #1
lastlaugh
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Any suggestions on how to integrate
[tex]\int cos^3xdx[/tex]
My first clue would be to substitute
(1-(Sinx)^2)cosxdx but any clue to go after this?
 
Last edited:
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  • #2
Well, do any of the techniques you've learned apply to

[tex]\int (1 - \sin^2 x) \cos x \, dx[/tex]

?
 
  • #3
Put sinx=t => cosx dx=dt

so u have int(1-t2)dt=t-t3/3
 
  • #4
thanks for the help. This trig integration stuff is killer

OK the reason I asked was because I have a lab assignment where I have to calculate the powers of [tex]\int cos^n[x]dx[/tex] where n is odd numbers from 3-13.
So can someone please look over my work for [tex]\int cos^5[x]dx[/tex] and make sure it looks valid:
First I substituted [tex](1-sin^2[x])^2[/tex] for [tex]cos^4[x][/tex] which gives me:
[tex]\int (1-sin^2[x])^2 cos[x]dx[/tex]
Then I let u=sinx meaning du=cosxdx
Giving me:
[tex]\int (1-u^2)^2du[/tex]
Which I simplfied to give me:
[tex]\int u^4 du - \int 2u^2 du + \int du[/tex]
Which gives me
[tex](u^5)/5-(2u^3)/3 + u[/tex]

Anyone see any glaring errors?
Thank you in advance.
 
Last edited:
  • #5
Originally posted by lastlaugh
thanks for the help. This trig integration stuff is killer

OK the reason I asked was because I have a lab assignment where I have to calculate the powers of [tex]\int cos^n[x]dx[/tex] where n is odd numbers from 3-13.
So can someone please look over my work for [tex]\int cos^5[x]dx[/tex] and make sure it looks valid:
First I substituted [tex](1-sin^2[x])^2[/tex] for [tex]cos^4[x][/tex] which gives me:
[tex]\int (1-sin^2[x])^2 cos[x]dx[/tex]
Then I let u=sinx meaning du=cosxdx
Giving me:
[tex]\int (1-u^2)^2du[/tex]
Which I simplfied to give me:
[tex]\int u^4 du - \int 2u^2 du + \int du[/tex]
Which gives me
[tex](u^5)/5-(2u^3)/3 + u[/tex]

Anyone see any glaring errors?
Thank you in advance.



WRONGGGGGGGGGGGGGGGGGGGG

is (1-u^2) du.
 
  • #6
Looks right to me. (except you should substitute back what u is)
 
  • #7
Originally posted by PrudensOptimus
WRONGGGGGGGGGGGGGGGGGGGG

is (1-u^2) du.

How it is wrong

Moreover General way will be to form a reduction formula
 
Last edited:
  • #8
Originally posted by Hurkyl
(except you should substitute back what u is)
Yeah I didn't feel like typing out any more latex:smile:
 

1. How do you integrate cos^3xdx?

To integrate cos^3xdx, you can use the substitution method by substituting u = sinx. This will change the integral to -sin^3x dx, which can be integrated using the power rule.

2. What is the benefit of using the substitution (1-(Sinx)^2)cosxdx in the integral?

The substitution (1-(sinx)^2)cosxdx helps in simplifying the integral by reducing the power of cosx from 3 to 1. This makes it easier to integrate using the power rule.

3. Can the substitution (1-(Sinx)^2)cosxdx be used in any integral involving cosx?

No, the substitution (1-(sinx)^2)cosxdx is specifically used for integrating cos^3xdx. It may not be applicable in other integrals involving cosx.

4. What is the purpose of substituting u = sinx in the integral?

Substituting u = sinx helps in simplifying the integral and making it easier to integrate. It also helps in reducing the power of cosx, making it more manageable to use the power rule.

5. Are there any other methods to integrate cos^3xdx?

Yes, there are other methods such as using trigonometric identities or integration by parts. However, the substitution method with u = sinx is the most efficient and commonly used method for integrating cos^3xdx.

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