Where can I find a demostration of the integral of sin(x)

[V12-POWER]

I know this is an identity, We have started with integrals but I can't figure it out why is the integral of sin(x) equal to -cos(x) + C.

Since integral calculus is antiderivative, yet the result of such integral is the very derivative of sin(x).

. . .sin(x) = -cos(x)

im confused with this. And when I try to solve it by substitution I don't end with the same result.
Thanks in advance

 

[JeffM]

I know this is an identity, We have started with integrals but I can't figure it out why is the integral of sin(x) equal to -cos(x) + C.

Since integral calculus is antiderivative, yet the result of such integral is the very derivative of sin(x).

. . .sin(x) = -cos(x)

im confused with this. And when I try to solve it by substitution I don't end with the same result.
Thanks in advance

\(\displaystyle f(x) = sin(x) \implies f'(x) = PLUS\ cos(x).\)

\(\displaystyle g(x) = cos(x) \implies g'(x) = MINUS\ sin(x).\)

I have no clue why you say \(\displaystyle sin(x) = cos(x).\) That is generally false.

We want to find a function the derivative of which is sin(x).

\(\displaystyle \text {Given } g(x) = cos(x) - C \text { and } h'(x) = sin(x).\)

\(\displaystyle \therefore g'(x) = -\ sin(x) = (-\ 1) * sin(x) = -\ h'(x) \implies\)

\(\displaystyle \displaystyle \int -\ h'(x)\ dx = \int g'(x)\ dx \implies -\ \int h'(x)\ dx = g(x) = cos(x) - C \implies\)

\(\displaystyle h(x) = -\ (cos(x) - C) = -\ cos(x) + C.\)

You need to keep track of the signs.

 

[stapel]

I know this is an identity

You know "what" is an identity?

We have started with integrals but I can't figure it out why is the integral of sin(x) equal to -cos(x) + C.

The integral is the anti-derivative. What do you get if you differentiate -cos(x) + C? If you get "sin(x)", then you've proven the equality, int(sin(x))dx = -cos(x) + C.

Since integral calculus is antiderivative, yet the result of such integral is the very derivative of sin(x).

. . .sin(x) = -cos(x)

The above is not an identity, as -tan(x) is not always equal to 1 (which is one result of your equation).

Did you perhaps mean to refer to the derivative of sine? So that you actually meant the following?

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sin(x)\, =\, -\cos(x)\)

Because, if so, this is not correct. Check your derivative tables. The correct derivative is:

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sin(x)\, =\, \cos(x)\)

So the integral of the sine function is not equal to the derivative of the sine function.

If I have misunderstood your question, kindly please reply with clarification. Thank you!

 

[V12-POWER]

You know "what" is an identity?


The integral is the anti-derivative. What do you get if you differentiate -cos(x) + C? If you get "sin(x)", then you've proven the equality, int(sin(x))dx = -cos(x) + C.


The above is not an identity, as -tan(x) is not always equal to 1 (which is one result of your equation).

Did you perhaps mean to refer to the derivative of sine? So that you actually meant the following?

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sin(x)\, =\, -\cos(x)\)

Because, if so, this is not correct. Check your derivative tables. The correct derivative is:

. . . . .\(\displaystyle \dfrac{d}{dx}\, \sin(x)\, =\, \cos(x)\)

So the integral of the sine function is not equal to the derivative of the sine function.

If I have misunderstood your question, kindly please reply with clarification. Thank you!

Exuse me because english is not my native language and maybe im using the wrong terms. Yes I had mixed some things, I thought the derivative of sin(x) was equal to -cos(x). The way trigonometric functions work is giving me a hard time.
Now, this is what im doing ending in a wrong result

[FONT=q_serif]∫ sin(x) dx

u= sin(x)
du= cos(x) dx
1/cos(x)du = dx
[/FONT][FONT=q_serif]
Then

∫u (1/cos(x)) du[/FONT][FONT=q_serif]

Im stuck here because I can't end with anything. Maybe im approaching the problem incorrectly, but the idea is to solve every integral by sustitution[/FONT]

 

[JeffM]

Exuse me because english is not my native language and maybe im using the wrong terms. Yes I had mixed some things, I thought the derivative of sin(x) was equal to -cos(x). The way trigonometric functions work is giving me a hard time.
Now, this is what im doing ending in a wrong result

[FONT=q_serif]∫ sin(x) dx

u= sin(x)
du= cos(x) dx
1/cos(x)du = dx
[/FONT][FONT=q_serif]
Then

∫u (1/cos(x)) du[/FONT][FONT=q_serif]

Im stuck here because I can't end with anything. Maybe im approaching the problem incorrectly, but the idea is to solve every integral by sustitution[/FONT]

Did you even bother to read my first post, which gave the demonstration that you initially asked for?

Your method goes nowhere because your integral has not fully replaced x by u.

\(\displaystyle u = sin(x) \implies \dfrac{du}{dx} = cos(x) = \sqrt{1 - sin^2(x)} = \sqrt{1 - u^2} \implies\)

\(\displaystyle dx = \dfrac{du}{\sqrt{1 - u^2}} \implies \displaystyle \int sin(x)\ dx = \int \dfrac{u}{\sqrt{1 - u^2}}\ du.\)

Now you have an integral that is completely in u. It's not the easiest to solve, but give it a go. Once you have done so, you will find that your proposed method works (although it is not easiest method).