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Integral of cos^3(x)

A specialized form of u-substitution involves taking advantage of trigonometric identities.

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  • leaf blue style avatar for user Jan Staněk
    Why not substitute "sinx = u" right at the beginning? Without having to distribute the brackets.
    From ∫cosx*(1-sin^2 x) dx you will get:
    ∫(1-u^2)du = ∫du - ∫u^2 du = u - (1/3)*u^3
    Substitute back and i got the same result:
    sinx-((1/3)*sin^3 x) + C
    (53 votes)
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  • piceratops seedling style avatar for user DatsyukFan
    Why doesnt u substitution work in this case in the original equation? let u = cosx
    u^3 's antiderrivative is u^4 / 4, then sub cosx back in?
    (8 votes)
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    • piceratops ultimate style avatar for user Just Keith
      That doesn't work because you do not have a du to go with the u.
      If u = cos x, then du = - sin x dx
      You don't have the - sin x, so you cannot make this substitution.

      Remember that in integrals, to use one of the standard forms, you need to have "du" which is the derivative of whatever you decide to call u. The "du" in the notation is not just a notational requirement, it really does have to be there or you don't get the correct answer.
      If the problem had been ∫ −cos³(x) sin(x) dx then that would be as you suggested ¼ cos⁴x + C
      (20 votes)
  • mr pink red style avatar for user Rainman
    intg [ 1/(1+cosx) ] dx ? how will we solve it?
    (5 votes)
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  • blobby green style avatar for user Jay Hansen
    Is there a reason that you can't use 1-sinx as u?
    (4 votes)
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    • leafers ultimate style avatar for user TripleB
      Well you have a 1 - sin^2 in the integral after separating the problem into cos * cos^2.
      You can try to make u = 1 - sin^2
      but I don't believe that would help.
      Then du = -2 sin cos
      and you don't have that in the expression.

      Hope I understood your question correctly.
      Let me know if you meant something else.
      (6 votes)
  • piceratops seed style avatar for user umar sayed
    what is the integration for sin^4 2x
    (1 vote)
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    • purple pi purple style avatar for user redthumb.liberty
      Time to break out the double angle identity substitutions:

      Let:

        f(x) = sin^4(2•x)

      F(x) = ∫( sin^4(2•x) )dx


      For reference:

      Angle Sum and Difference Identities:

       sin(α + β) = sin(α)•cos(β) + cos(α)•sin(β)

      sin(α – β) = sin(α)•cos(β) – cos(α)•sin(β)

      cos(α + β) = cos(α)•cos(β) – sin(α)•sin(β)

      cos(α – β) = cos(α)•cos(β) + sin(α)•sin(β)

      tan(α + β) = (tan(α) + tan(β)) ÷ (1 - tan(α)•tan(β))

      tan(α - β) = (tan(α) - tan(β)) ÷ (1 + tan(α)•tan(β))


      Decompose trig^n(x) to equivalent trig^2(x) multiplications:

        F(x) = ∫( sin^2(2•x)•sin^2(2•x) )dx


      Evaluate sub:

          u = 2•x

      ( d/dx(2•x) )dx = du

      ( 2•d/dx(x) )dx = du

      ( 2•dx/dx )dx = du

      ( 2 )dx = du

      dx = ( 1/2 )du


      Input sub:

        F(u) = ∫( sin^2(u)•1/2•sin^2(u)•1/2 )du

      F(u) = 1/4•∫( sin^2(u)•sin^2(u) )du


      Find sum angle idendity for sin^2(x) if α = β:

          cos(α + β) = cos(α)•cos(β) – >sin(α)•sin(β)<

      α = β = θ

      cos(θ + θ) = cos(θ)•cos(θ) – sin(θ)•sin(θ)

      cos(2•θ) = cos^2(θ) – sin^2(θ)

      sin^2(θ) + cos(2•θ) = cos^2(θ)

      sin^2(θ) = cos^2(θ) - cos(2•θ)


      Isolate equivalent for cos^2(θ) from pythagorean inedtity:

            sin^2(θ) + cos^2(θ) = 1

      cos^2(θ) = 1 - sin^2(θ)

      sin^2(θ) = 1 - sin^2(θ) - cos(2•θ)

      sin^2(θ) + sin^2(θ) = 1 - cos(2•θ)

      2•sin^2(θ) = 1 - cos(2•θ)

      sin^2(θ) = 1/2 - cos(2•θ)/2

      u = θ

      sin^2(u) = 1/2 - cos(2•u)/2


      Input trig equivalent:

        F(u) = 1/4•∫( (1/2 - cos(2•u)/2)•(1/2 - cos(2•u)/2) )du

      F(u) = 1/8•∫( (1 - cos(2•u))•(1 - cos(2•u)) )du

      F(u) = 1/8•∫( 1 - 2•cos(2•u) + cos^2(2•u) )du

      F(u) = 1/8•( ∫(1)du - ∫(2•cos(2•u))du + ∫(cos^2(2•u))du )


      Evaluate sub:

          v = 2•u

      ( d/du(2•u) )du = dv

      ( 2•d/dx(u) )du = dv

      ( 2•du/du )du = dv

      ( 2 )du = dv

      du = ( 1/2 )dv


      Input sub:

        F( ) = 1/8•( ∫(1)du - ∫(2•cos(v)•1/2)dv + ∫(cos^2(v)•1/2)dv )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/2•∫(cos^2(v))dv )


      Find sum angle idendity for cos^2(x) if α = β:

          cos(α + β) = >cos(α)•cos(β)< – sin(α)•sin(β)

      α = β = θ

      cos(θ + θ) = cos(θ)•cos(θ) – sin(θ)•sin(θ)

      cos(2•θ) = cos^2(θ) – sin^2(θ)

      cos^2(θ) – sin^2(θ) = cos(2•θ)

      cos^2(θ) = cos(2•θ) + sin^2(θ)


      Isolate equivalent for sin^2(θ) from pythagorean inedtity:

            sin^2(θ) + cos^2(θ) = 1

      sin^2(θ) = 1 - cos^2(θ)

      cos^2(θ) = cos(2•θ) + 1 - cos^2(θ)

      cos^2(θ) + cos^2(θ) = cos(2•θ) + 1

      2•cos^2(θ) = cos(2•θ) + 1

      cos^2(θ) = cos(2•θ)/2 + 1/2

      v = θ

      cos^2(v) = cos(2•v)/2 + 1/2


      Input trig equivalent:

        F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/2•∫( cos(2•v)/2 + 1/2 )dv )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•∫( cos(2•v) + 1 )dv )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(2•v))dv + ∫(1)dv ) )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(2•v))dv + v ) )


      Evaluate sub:

          w = 2•v

      ( d/dv(2•v) )dv = dw

      ( 2•d/dv(v) )dv = dw

      ( 2•dv/dv )dv = dw

      ( 2 )dv = dw

      dv = ( 1/2 )dw


      Input sub:

        F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( ∫(cos(w)•1/2)dv + v ) )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( 1/2•∫(cos(w))dv + v ) )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/4•( 1/2•sin(w) + v ) )

      F( ) = 1/8•( ∫(1)du - ∫(cos(v))dv + 1/8•sin(w) + v/4 )

      F( ) = 1/8•( ∫(1)du - sin(v) + 1/8•sin(w) + v/4 )

      F( ) = 1/8•( u - sin(v) + 1/8•sin(w) + v/4 )

      F( ) = u/8 - sin(v)/8 + 1/64•sin(w) + v/32


      Back sub w → v:

          w = 2•v

      F( ) = u/8 - sin(v)/8 + 1/64•sin(2•v) + v/32


      Back sub v → u:

          v = 2•u

      F(u) = u/8 - sin(2•u)/8 + 1/64•sin(2•2•u) + 2•u/32

      F(u) = u/8 - sin(2•u)/8 + 1/64•sin(4•u) + u/16

      F(u) = 2•u/16 + u/16 - sin(2•u)/8 + 1/64•sin(4•u)

      F(u) = 3•u/16 - sin(2•u)/8 + 1/64•sin(4•u)


      Back sub u → x:

          u = 2•x

      F(x) = 3•2•x/16 - sin(2•2•x)/8 + 1/64•sin(4•2•x)

      F(x) = [3•x/8 - sin(4•x)/8 + sin(8•x)/64 + C]
      (10 votes)
  • blobby green style avatar for user aantoniev97
    How can we solve for example cos^7x ?
    (6 votes)
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  • leaf green style avatar for user V R  Sri
    Can you integrate the log of a trig function, such as log (sin x), or log cos x, without the provision of "limits".
    Or does the solution necessarily require "limits", such as classic textbook problem " integration of log(sin x).dx with limits from 0 to (pi/2)"
    (3 votes)
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  • piceratops tree style avatar for user jgar95
    When I tried on my own I got this:
    ∫cos³(x)dx = ∫ cos(x) - ∫cos(x) sin(x))ˆ2 dx
    Then I decided to work on ∫cos(x) sin(x))ˆ2 dx using integration by parts:
    > ∫cos(x) sin(x))ˆ2 dx = (sin(x))ˆ2cos(x)-2∫cos(x) sin(x))ˆ2dx
    > 3∫cos(x) sin(x))ˆ2 dx = (sin(x))ˆ2cos(x)
    > ∫cos(x) sin(x))ˆ2 dx = [ (sin(x))ˆ2cos(x) ]/3
    But when i substitute it at the original formula doesn't seem right.
    Where is the error?
    (1 vote)
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  • old spice man green style avatar for user Dania  Zaheer
    2(1-cos2x) will be = 2-2cos2x or 2-cos4x ? Can someone please answer this ASAP?
    (1 vote)
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  • piceratops seed style avatar for user Shuhat Chowdhury
    hi, this might be a dumb question, but isn't (cos x) (cos^2x) be cos^3 x^3
    (1 vote)
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Video transcript

- [Voiceover] Let's see if we can take the indefinite integral of cosine of X to the third power. I encourage you to pause the video and see if you can figure this out on your own. You have given it a go and you might have gotten stuck. Some of you all might have been able to figure it out, but some of you all might have gotten stuck. You're like, "Okay, cosine to the third power. "Well, gee, if I only had a derivative of cosine here, "if I had a negative sign of X or a sin of X here, "maybe I could've used U substitution, "but how do I take the anti-derivative "of cosine of X to the third power?" The key here is, is to use some basic trigonometric identities. What do I mean by that? We know that sin squared X plus cosine squared X is equal to one, or if we subtract sin squared from both sides, we know that cosine squared X is equal to one, write it this way, is equal to one minus sin squared X. What would happen if cosine to the third power, that's cosine squared times cosine. What happens if we were to take that cosine squared? Let me just rewrite it. This is the same thing as cosine of X times cosine squared of X, DX. What if we were to take this thing right over here, let me do that magenta color. What if we were to take this right over here and replace it with this. I now what you're thinking. "Sal, what's that going to do for me? "This feels like I'm making this integral "even more convoluted." What I would tell you, I would say, "This might seem like it's getting more complicated, "but as you explore and you play with it, "you'll see that this actually makes "the integral more solvable." Let's try it out. If we do that, this is going to be equal to the indefinite integral, cosine of X times one minus sin squared X, DX. What is this going to be equal to? This is going to be equal to, let me do this in that green color. This is going to be equal to the indefinite integral of cosine X. I'm just going to distribute the cosine of X. Cosine of X minus, minus cosine of X, cosign of X sin squared of X, sin squared, sin squared X and then I can close the parentheses, DX. This, of course, is going to be equal to the integral of cosine of X, DX, and we know what that's going to be, minus the integral. I'll switch to one color now, of cosine of X, sin squared X, sin squared X, DX. Now, this is where it gets interesting. This part right over here is pretty straight forward. The anti-derivative of cosine of X is just sin of X. This right over here is going to be sin of X. I'll worry about the plus C at the end because both of these are going to have a plus C, so might as well just put one big plus c at the end. That's sin of X and then what do we have going on over here? Well, you might recognize, I have a function of sin of X. I'm taking sin of X and I'm squaring it and then I have sin of X's derivative right over here. This fits the, I have some derivative of a function and then I have another and then I have a, I guess you could say, a function of that function. G of F of X. That's a sin that maybe U substitution is in order, or we've seen the pattern, we've seen this show multiple times already, that you could just say, "Okay, if I have a function of a function "and I have that functions derivative, "then essentially I can just take the anti-derivative "with respect to this function." This would be equal to, say, capital G is the anti-derivative of lower case G. Capital G of F of X plus C. Now, if what I said didn't make sense, then we could do U substitution and go through it a little bit more step by step. Let's just do that because we want things to make sense. That's the whole point of these videos. We could say U is equal to sin of X and then DU is going to be equal to cosine of X, DX. This part and that part is going to be DU and then this is going to be U squared. This is going to be minus. We have the integral of U squared, DU. What is this going to be? This is going to be, we're going to have negative U to the third power over three. Then, we know what U is. The U is equal to sin of X. We have our sin of X here for the first part of the integral, for the first integral. We have the sin of X and then this is going to be minus. Let me just write it this way. Minus 1/3 minus 1/3. Instead of U to the third, we know U is sin of X. Sin of X to the third power. Then now, we can throw that plus C there. We're done. We've just evaluated that indefinite integral. The key to it is to just play around a little bit with trigonometric identities so that you can get the integral to a point that you can use the reverse chain rule or you can use U substitution, which is just really another way of expressing the reverse chain rule.