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### Course: Calculus, all content (2017 edition) > Unit 4

Lesson 3: Indefinite integrals of common functions- Indefinite integrals of sin(x), cos(x), and eˣ
- Indefinite integral of 1/x
- Indefinite integrals: eˣ & 1/x
- Particular solutions to differential equations: rational function
- Particular solutions to differential equations: exponential function
- Particular solutions to differential equations
- Indefinite integrals: sin & cos
- Integrating trig functions
- Antiderivatives and indefinite integrals review
- Common integrals review

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

∫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.

## Want to join the conversation?

- 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(77 votes)- Short answer: yes.

You could add a constant on each term, but since they're arbitrary, they can all be added together into a single, arbitrary constant.(174 votes)

- You said that we learned about the natural log in the last video, which video are you talking about?(53 votes)
- It's actually in the next video, the Antiderivative of x^-1.(67 votes)

- at3:33you said always remember about constant "c",its important . why it is?(15 votes)
- 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)

- what is difference between indefinite integral and definite integral?(15 votes)
- the indefinate integral isn't between two set values and the definate integral is(27 votes)

- why is it the natural log of the absolute value of a and not ln(a)?(15 votes)
- Due to the fact that the result of taking the ln of a negative value is undefined. The absolute value application allows negative values to be defined as well since it turns negative values into positive. In conclusion the absolute value application gives the anti-derivative the same domain as 1/x.(11 votes)

- 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?(5 votes)
- 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.(12 votes)

- isn't the derivative of sin(t) = -cos(t)?(3 votes)
- The anti-derivative of sin(t) is -cos(t) but the derivative of sin(t) is cos(t)(9 votes)

- 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)
- 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)
- It is not; adding any constant to

furnishes yet another antiderivative of*-cos*

. There are in fact infinitely many functions whose derivative is*sin*

.*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.)(8 votes)

- Life advice : Always remember the constant!!(4 votes)
- huh yeah. that can come back and sting u if u forget it(2 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.