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Integral of sin^2(x) cos^3(x)

Another example where u substitution combined with certain trigonometric identities can be used.

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  • blobby green style avatar for user Nouman
    I think I'm much more comfortable with the u-substitution as compared to the REVERSE CHAIN RULE. Will that be a problem or is it alright?
    (14 votes)
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  • blobby green style avatar for user theonlypersonhere
    I got a different answer after substituting the trig identity for sin(x)^2 = 1 - cos(x)^2
    So what I have now is ∫(1 - cos(x)^2) * cos(x)^3 * dx
    Then after distributing the cos(x)^3 I have, ∫cos(x)^3 - ∫cos(x)^5 * dx
    Evaluating this give me: (cos(x)^4)/4 - (cos(x)^6)/6
    Is this equivalent to the answer in the video, or where have I gone wrong?
    (1 vote)
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    • leaf orange style avatar for user nicklaus
      Basically you can't integrate the cos(x)^3 and the cos(x)^5. The reason for this is because they represent cos(x)*cos(x)*cos(x) and you cant integrate that without the u substitution. So in the end your only choice is to sub the trig identity for the cos(x)^2
      (8 votes)
  • blobby green style avatar for user Manoj Cracked
    Please solve ∫(2-sin2x/1-cos2x)e^x
    (1 vote)
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    • leaf grey style avatar for user Qeeko
      I will assume you intend the integrand to be interpreted as [ (2 - sin 2x) / (1 - cos 2x) ]eᵡ. To solve the integral, we will first rewrite the sine and cosine terms as follows:

      I) sin(2x) = 2sin(x)cos(x);
      II) cos(2x) = 2cos²(x) - 1.

      Rewriting yields
      2 - sin(2x)
      = 2 - 2sin(x)cos(x)
      = 2[1 - sin(x)cos(x)],

      and
      1 - cos(2x)
      = 1 - [2cos²(x) - 1]
      = 2 - 2cos²(x)
      = 2[1 - cos²(x)]
      = 2sin²(x).

      Hence
      [ (2 - sin 2x) / (1 - cos 2x) ]eᵡ
      = ( 2[1 - sin(x)cos(x)] / [ 2sin²(x) ] )eᵡ
      = [ 1/sin²(x) - cos(x)/sin(x) ]eᵡ
      = eᵡ / sin²(x) - eᵡcot(x).

      Thus ∫ [ (2 - sin 2x) / (1 - cos 2x) ]eᵡ dx = ∫ [ eᵡ / sin²(x) - eᵡcot(x) ] dx. This may be split up into two integrals as ∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx. We will first focus on the first of these integrals.

      Recall that d/dx cot(x) = -1 / sin²(x). Using integration by parts on the expression ∫ eᵡ / sin²(x) dx yields

      ∫ eᵡ / sin²(x) dx
      = -eᵡcot(x) + ∫ eᵡcot(x) dx.

      When we plug this into the expression ∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx, we get

      ∫ eᵡ / sin²(x) dx - ∫ eᵡcot(x) dx
      = -eᵡcot(x) + ∫ eᵡcot(x) dx - ∫ eᵡcot(x) dx
      = -eᵡcot(x) + C.

      Therefore, the answer is -eᵡcot(x) + C, where C is an arbitrary real number.
      (7 votes)
  • leaf orange style avatar for user Lex
    Can anyone help me with this: ∫ ln(x)/sin(x) dx
    (3 votes)
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  • blobby green style avatar for user izabelebasi
    Where does the du go after we solve for the anti derivative of U?
    (2 votes)
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  • leaf green style avatar for user Eric Parker
    why can't ∫sin^2(x)*cos^2(x)*cos(x) be written as ∫1*cos(x)?
    after all trig identity says sin^2(x)*cos^2(x)=1
    (2 votes)
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  • duskpin ultimate style avatar for user staceyrivet
    Is it possible to use U substitution without distributing cosx to (1- sinx^2)? I didn't distribute cos(x) and therefore my solution was different: x - 1/3(sin^3(x)) + C instead of sin(x) - 1/3(sin^3(x)) + C. I feel that when I do distribute my professor does not and ends up finding a different solution than what I would have done and then when I dont distribute, he does. Is there a rule I am missing here?
    (2 votes)
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    • hopper cool style avatar for user Oliver Dahl
      In the beginning of learning a concept (like U-sub), it is a good idea to exaggerate the amount of steps it takes to form a solution.

      Some little things might be lost if we jump too far between the different steps.

      It might be a good idea to control the solutions by deriving the finished antiderivative.

      (x - 1/3(sin^3(x)) + C)'=cos^3(x)-cos(x)+1
      (sin(x) - 1/3(sin^3(x)) + C)'=cos^3(x)

      What could we do to make these derivatives equal eachother?

      I hope this was a little helpful!
      (1 vote)
  • blobby green style avatar for user Mike Bell
    How do we know not to do sin^2(x)-sin^4(x)=-sin^2(x)?
    (1 vote)
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  • blobby green style avatar for user erretreinta
    Hello, thanks a lot for the video!
    I would like to know why does Sal distributes this way ():
    sin² (1 - sin²)
    instead of:
    (1 - sin²) cos
    Is it an intuitive thing to do or is there any kind of hint for doing this?
    Cheers :)
    (1 vote)
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  • aqualine ultimate style avatar for user TEO JIA MING
    let y= (cosx)^4
    dy/dx = 4 (cos x) ^3 * (sin x)^2

    so my answer is 1/4 { cosx) ^4 } + C, is that correct?
    (0 votes)
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Video transcript

- [Voiceover] Let's see if we can take the indefinite integral of sine squared x cosine to the third x dx. Like always, pause the video and see if you can work it through on your own. All right, so right when you look at it you're like, "Oh wow, if this was just a sine of x, "not a sine squared of x. "Well that's going to be the negative of "the derivative of cosine of x. "Maybe I could have used u-substitution. "Likewise, if this was just a cosine of x not a cosine "to the third of x, I could have used u-substitution. "I could have said u is equal to sine of x, "but I can't do that over here." So the general ideas here, if one of these has an odd exponent and you see that this one does have an odd exponent, the cosine has an odd exponent. What you do is you try to algebraically engineer this expression inside, so that you can use u-substitution. The way that you can do that, if you have an odd exponent like this, is to separate out one of the cosine's x and then use the Pythagorean Identity with the remaining cosine squared x, what do I mean by that? Let me just rewrite this, so this can be rewritten as sine squared x times... Let me write it this way. Times cosine squared x cosine x. All I did is I rewrote sometime to the third power, something to the second power times that thing to the first power, dx. And this could be rewritten as, sine squared x, then I'm going to use the Pythagorean Identity to convert this into one minus sine squared. So that, by the Pythagorean Identity, is the same thing as one minus sine squared x and then we have cosine x, times cosine x dx. Now I can distribute this sine squared times one minus sine squared and I am left with... Let me do this in a new color, this business right here. Now I'm left with the indefinite integral of, Sine squared x times one is going to be sine squared x and then sine squared x times negative sine squared x is negative sine to the fourth. Then all of that times cosine x. All of that times cosine x dx. Now this is starting to look interesting, cause I have sine squared x minus sine to the fourth x, but I have the derivative of sine sitting out here. I have cosine x, that's the whole reason why we did this little algebraic manipulation. So u-substitution works out quite well now. Because if we said that.. Let me do another color, I'll do purple. If we say that u is equal to sine of x, then du is going to be equal to cosine of x dx, and that works out quite well because we have the du right over here. Then this would be u squared minus u to the fourth. Which we know how to take the anti-derivative of, it's over the whole stretch. We can rewrite this as the indefinite integral. Instead of sine squared x we're saying sine is the same thing as u so we can rewrite that as, u squared minus u to the fourth times du. This is pretty straight forward, now this is going to be u to the third over three minus u to the fifth over five plus c. Then we just do the reverse substitution and then that gets us to be, instead of u we want to put a sine of x there. So we're gonna get sine to the third, sine of x to the third over three minus sine of x to the fifth power. Let me rewrite that five a little bit. Sine of x to the fifth power over five plus c and we are done.