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Indefinite integrals of sin(x), cos(x), and eˣ

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)
∫sin(x)dx=-cos(x)+C, ∫cos(x)dx=sin(x)+C, and ∫eˣdx=eˣ+C. Learn why this is so and see worked examples. Created by Sal Khan.

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  • mr pants teal style avatar for user Nathan Fernandes
    It might be a silly question, but, I don't have to put a constant to each operation?

    I saw that he correct himself in the video, but, he put just one constant for two operations, or that C is something like A + B ( the A for the sinx anti-derivative and the B for the cosx anti-derivative) ?

    Thanks in advance
    (76 votes)
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  • blobby green style avatar for user briana collins
    You said that we learned about the natural log in the last video, which video are you talking about?
    (53 votes)
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  • leaf green style avatar for user shariar
    at you said always remember about constant "c",its important . why it is?
    (15 votes)
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    • leaf orange style avatar for user ↑←xylon97→↓
      Consider ∫ 1/2x dx, without using the constant integration.

      Method 1
      ∫ 1/2x dx
      = 0.5 ∫ 1/x dx
      = 0.5 ln(x) ...(1)

      Method 2
      Let u = 2x => du/dx = 2 => dx = du/2
      ∫ 1/2x dx
      = ∫ 1/2u du
      = 0.5 ∫ 1/u du
      = 0.5 ln(u)
      = 0.5 ln(2x) ...(2)

      From (1) we have ∫ 1/2x dx = 0.5 ln(x).
      From (2) we have ∫ 1/2x dx = 0.5 ln(2x).

      Conclusion: x = 2x?
      Not quite.

      Where did we go wrong? The +C.
      The results can be explained by using a property of logarithms-
      ln(ab) = ln(a) + ln(b).
      So 0.5 ln(2x) = 0.5[ln(2) + ln(x)] = 0.5 ln(x) + 0.5 ln(2) = 0.5 ln(x) + C.
      This is why the +C is very important.
      (65 votes)
  • piceratops ultimate style avatar for user Min ChanHong
    what is difference between indefinite integral and definite integral?
    (13 votes)
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  • male robot hal style avatar for user tom gibbs
    isn't the derivative of sin(t) = -cos(t)?
    (3 votes)
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  • male robot hal style avatar for user Rohan Khajuria
    What is e? It always pops up in the mathematical world, but I can never figure out what it is. Has Khan made any videos on it?
    (3 votes)
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    • male robot hal style avatar for user bruno.garcia
      It actually pops up all over the natural world. It's a somewhat mysterious constant called Euler's number (~ 2.718). It's also the implicit base for natural logarithm (Ln) and useful due to its properties. Yes, Khan made videos on it, look for compound interest and e.
      (10 votes)
  • blobby green style avatar for user Pierre-Olivier
    What is the antiderivative of e^(4x)? None of sal's examples for antiderivatives include chain rule stuff. I read everywhere that the antiderivative of e^(4x) is e^(4x)/4. It makes sense, because if you asked me to find the derivative of e^(4x)/4, I would do the chain rule by multiplying that by 4 (which is the derivative of 4x), which would give me 4e^(4x)/4, equaling the original e^(4x). But I don't understand how to get back there with the antiderivative.
    (4 votes)
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  • female robot ada style avatar for user Raviv Sarch
    It's pretty easy to see that the derivative of -cosx is sinx, but how can you prove that the only anti derivative of sinx is -cosx?
    (2 votes)
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    • leaf grey style avatar for user Qeeko
      It is not; adding any constant to -cos furnishes yet another antiderivative of sin. There are in fact infinitely many functions whose derivative is sin.

      To prove that two antiderivatives of a function may only differ by a constant, follow this outline: suppose a function ƒ has antiderivatives F and G. Define a function H by H = F - G. Conclude that H' = 0, so that H is a constant; F - G = C holds for some constant C. Thus F = G + C. It is not hard to make this "proof" rigorous, and I suggest you do so.

      (Note: when we conclude from the fact that H' is zero that H is constant, we actually use the mean value theorem.)
      (7 votes)
  • blobby green style avatar for user mira_ocean_sky
    what is the natural log? in which video ?? I am so confused . I have so many missing stuff!
    (1 vote)
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  • blobby green style avatar for user aashifrazeen
    why is the integration of. Sin(2x -3) = (-1/2)cos (3-2x ) ; here why is it (3-2x ) instead of (2x -3) ?
    (3 votes)
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    • leafers ultimate style avatar for user TripleB
      Good question.
      Remember from Trig class that:
      Cos (x) = Cos (-x)
      You can visually see this on the unit circle that any point you pick on the circle will have the same Cos value as it's negative angle.

      So then
      Cos (2x - 3) = Cos ( - (2x - 3)) = Cos (3 - 2x)
      So you could show the answer either way and they are equivalent.
      (3 votes)

Video transcript

I thought I would do a few more examples of taking antiderivatives, just so we feel comfortable taking antiderivatives of all of the basic functions that we know how to take the derivatives of. And on top of that, I just want to make it clear that it doesn't always have to be functions of x. Here we have a function of t, and we're taking the antiderivative with respect to t. And so you would not write a dx here. That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative, of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. The derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t-- so plus sine of t. And we're done. We've found the antiderivative. Now let's tackle this. Now we don't have a t. We're taking the indefinite integral with respect to-- actually, this is a mistake. This should be with respect to a. Let me clean this up. This should be a da. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be da. That's what we are integrating or taking the antiderivative with respect to. So what is this going to be equal to? Well once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a da, so this one right over here-- a d I'll do it in green-- plus the indefinite integral, or the antiderivative, of 1/a da. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e in the exponential function in general is so amazing. And if we just replaced a with x or x with a, you get the derivative with respect to a of e to the a is equal to e to the a. So the antiderivative here, the derivative of e to the a, the antiderivative is going to be e to the a. And maybe you can shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me-- always important. Remember the constant. So you have a constant factor right over here. Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1/a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative, so there could be a constant factor out here as well. And we are done. We found the antiderivative of both of these expressions.