Simple Trig Integration

[Jason76]

$\displaystyle \int \cos(x) \sin(x)dx$

The answer is

$\displaystyle \frac{1}{2}\sin^{2}(x)$

Some type of substituting is going on. I do see that they used the intergal power rule on either cos or sin.

Any hints on the details?

 

[dsk2]

Hey Jason76,

Well, if you know $\displaystyle \int \sin(x)dx=-cos(x)+C$, you are mostly set. It would help if you also know that $\displaystyle 2sin(x)cos(x)=sin(2x)$


Now, $\displaystyle \int \cos(x) \sin(x)dx=\frac{1}{2}\int sin(2x) dx$. With me?
$\displaystyle \frac{1}{2}\int sin(2x) dx= -\frac{1}{4}cos(2x) +C$

That is your answer. You can use the identity $\displaystyle cos (2x)=1-2sin^2(x)$ to obtain the form that you desire. Note that when you add two constants, you end up with a new constant, which is still a constant.


Cheers,
Sai.

 

[soroban]

Hello, Jason76!

$\displaystyle \displaystyle\int \cos x \sin x\,dx$

$\displaystyle \text{Answer: }\:\tfrac{1}{2}\sin^2\!x+C$

We have: .$\displaystyle \displaystyle \int \sin x\,(\cos x\,dx)$

Let $\displaystyle u = \sin x \quad\Rightarrow\quad du = \cos x\,dx$

Substitute: .$\displaystyle \displaystyle \int u\,du \;=\;\tfrac{1}{2}\!u^2+C$

Back-substitute: .$\displaystyle \frac{1}{2}\sin^2\!x+C$

 

[Jason76]

Hello, Jason76!


We have: .$\displaystyle \displaystyle \int \sin x\,(\cos x\,dx)$

Let $\displaystyle u = \sin x \quad\Rightarrow\quad du = \cos x\,dx$

Substitute: .$\displaystyle \displaystyle \int u\,du \;=\;\tfrac{1}{2}\!u^2+C$

Back-substitute: .$\displaystyle \frac{1}{2}\sin^2\!x+C$

I had some problem understanding integration. But now corrected :

Function to Derviative thru Integration to Function (Example):

$\displaystyle sin^2x$

$\displaystyle y' = sin^2x$

Use the derivative power rule and (d)(du) - chain rule

$\displaystyle \sin x \cos x dx$

$\displaystyle \int \sin x \cos x dx$

Using intergal power rule and ignoring the $\displaystyle \cos$

$\displaystyle \dfrac{1}{2} \sin x + C$

But. I think I did something wrong because the original function should be $\displaystyle \dfrac{1}{2} \sin x$ shouldn't it?

 

[HallsofIvy]

I have no clue what you are writing or what you think you are doing.

If you are saying that you are given $\displaystyle y'= sin^2(x)$ and you want to find y, then, yes, you need to integrate (find the "anti-derivative") but it is NOT any thing you give there. The example soroban gave was integrating $\displaystyle \int sin(x)cos(x)dx$ which he did by "substitution", letting u= sin(x) so that du= cos(x)dx and the integral becomes $\displaystyle \int u du= \frac{1}{2}u^2+ C= \frac{1}{2}sin^2(x)+ C$.

Now, if you want to differentiate that , that is you have $\displaystyle y(x)= \frac{1}{2}sin^2(x)+ C$ and you want to find y', then you would use the "chain rule" which is essentially the reverse of using substitution to integrate: with $\displaystyle u(x)= sin(x)$ we have $\displaystyle y=\frac{1}{2} u^2$ so that $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{2}(2u)(cos(x))= sin(x)cos(x)$, just what you started with.

But to integrate $\displaystyle sin^2(x)$ you will need to use a trig identity: $\displaystyle sin^(x)= \frac{1}{2}(1- cos(2x))$ so that $\displaystyle \int sin^2(x)= \frac{1}{2}\int (1- cos(2x))x= \frac{1}{2}(x+ \frac{1}{2}sin(2x)))+ C$.

 

[JeffM]

I had some problem understanding integration. But now corrected :
I THINK I follow what you are thinking

You look at $\displaystyle \int cos(x) * sin(x)dx = \frac{1}{2}sin^2(x) + C$, and ask what is going on and work backward. That method

will explain answers, but it will not help you find answers.

Function to Derviative thru Integration to Function (Example):

$\displaystyle sin^2x$ But this not your answer of $\displaystyle \ \frac{1}{2}sin^2(x) + C.$

$\displaystyle y' = sin^2x$ This is wrong if you are indeed working backwards. You want to find the derivative of your answer.

Use the derivative power rule and (d)(du) - chain rule

$\displaystyle Let\ u = sin(x) \implies \dfrac{du}{dx} = cos(x)\ and\ F(x) = \frac{1}{2}sin^2(x) + C = \frac{1}{2}u^2 + C \implies \dfrac{d(F(x))}{du} = \frac{1}{2}(2u) + 0 = u = sin(x) \implies$

$\displaystyle \dfrac{d(F(x))}{dx} = \dfrac{d(F(x))}{du} * \dfrac{du}{dx} = sin(x) * cos(x) = cos(x) * sin(x).$

That shows you that the answer is correct. When you differentiate your answer you get what was to be integrated. But it does not help you do the integration in the first place.

$\displaystyle \sin x \cos x dx$

$\displaystyle \int \sin x \cos x dx$

Using intergal power rule and ignoring the $\displaystyle \cos$ You can't just ignore things

$\displaystyle \dfrac{1}{2} \sin x + C$

But. I think I did something wrong because the original function should be $\displaystyle \dfrac{1}{2} \sin x$ shouldn't it?

You want to integrate $\displaystyle \int cos(x) * sin(x)dx.$

You have to "SEE" that the product being integrated can be split into a function and its derivative. This is what is hard about integral calculus. Before you know which of many techniques to use, you have to recognize a pattern. Integral calculus has mechanical rules, but it requires pattern recognition as well. Splitting a product into a function and its derivative is one very common pattern.

$\displaystyle \int cos(x) * sin(x)\ dx = \int sin(x) * cos(x)\ dx.$ But cos(x) is the derivative of sin x, which suggests the substitution

$\displaystyle v = sin(x) \implies \dfrac{dv}{dx} = cos(x) \implies dv = cos(x)\ dx \implies \int sin(x) * cos(x)\ dx = \int u\ du = \frac{1}{2}u^2 + C = \frac{1}{2}sin^2(x) + C.$

 

[Jason76]

You want to integrate $\displaystyle \int cos(x) * sin(x)dx.$

You have to "SEE" that the product being integrated can be split into a function and its derivative. This is what is hard about integral calculus. Before you know which of many techniques to use, you have to recognize a pattern. Integral calculus has mechanical rules, but it requires pattern recognition as well. Splitting a product into a function and its derivative is one very common pattern.

$\displaystyle \int cos(x) * sin(x)\ dx = \int sin(x) * cos(x)\ dx.$ But cos(x) is the derivative of sin x, which suggests the substitution

$\displaystyle v = sin(x) \implies \dfrac{dv}{dx} = cos(x) \implies dv = cos(x)\ dx \implies \int sin(x) * cos(x)\ dx = \int u\ du = \frac{1}{2}u^2 + C = \frac{1}{2}sin^2(x) + C.$

(I'm thinking) Integration gets rid of a deriviative and leads to a family of functions, but it doesn't lead to a specific function. So integration on a derivative won't lead to the specific original function that was derived way back (before you took the derivative). But anyhow, I'm getting the right answer, despite any deep thinking, so that's all that matters.

 

[Deleted member 4993]

$\displaystyle \int \cos(x) \sin(x)dx$

The answer is

$\displaystyle \frac{1}{2}\sin^{2}(x)$ → That is incorrect - it should be → $\displaystyle \frac{1}{2}sin^{2}(x) + C$

Some type of substituting is going on. I do see that they used the intergal power rule on either cos or sin.

Any hints on the details?

.

 

[dsk2]

Jason,


You know you can preview your post before posting it. That way you can double check your superscripts and constants before posting your reply. Its easier when you are understood the way you want to be understood.

But nice thought process there. Yes, you made a mistake in your example.In your example the original function should not be $\displaystyle \frac{1}{2}sin^2(x)$. It should be $\displaystyle sin^2(x)$ as expected by you. The reason you got $\displaystyle \frac{1}{2}sin^2(x)$ is because you missed the coefficient 2 in your differentiation of $\displaystyle sin^2(x)$.

Usually, when you differentiate a function and integrate it back again, you get back the original form except for a translation. Hence the unknown constant. If you did it correctly and if you still do not see the original form of the function, it means it is an equivalent form and usually you can use some basic trig identities to transform it into the original function.

The family of functions that you describe is true as long as all they are look similar (graphically) or only differ by a constant mathematically. When you multiply by 1/2 you are scaling a function (think graphically-it shrinks to half the size). But when you lets say add 1/2 you are only pushing it up. The form of the original function hasn't changed by translation, but has changed by multiplication (with anything other than 1 of course).

Cheers,
Sai.

(I'm thinking) Integration gets rid of a deriviative and leads to a family of functions, but it doesn't lead to a specific function. So integration on a derivative won't lead to the specific original function that was derived way back (before you took the derivative). But anyhow, I'm getting the right answer, despite any deep thinking, so that's all that matters.

 

[srmichael]

$\displaystyle \int \cos(x) \sin(x)dx$

The answer is

$\displaystyle \frac{1}{2}\sin^{2}(x)$

Some type of substituting is going on. I do see that they used the intergal power rule on either cos or sin.

Any hints on the details?

Also, realize that $\displaystyle \int\cos(x)\sin(x)dx$ also equals $\displaystyle -\frac{1}{2}\cos^2(x)+C$ had we chose $\displaystyle u=\cos(x)$ instead of $\displaystyle u=\sin(x)$. This is good to know should this have been a multiple choice problem and the answer of $\displaystyle \frac{1}{2}\sin^2(x)+C$ was not an option.

 

[Deleted member 4993]

So now you have:

$\displaystyle \int sin(x) cos(x) dx \ =$

.............. .............. ½ sin²x + C₁ ....................... OR

........................ - ½ cos²x + C₂ ......................... OR

........................ - ¼ cos(2x) + C₃

All these functions differ only by constant value.......................