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Calculation of the integral of sine to power 4
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SDB2001

Membre
Posté le 06/11/2021 à 17:45:53

de SDB2001
Membre
Hello,

I need help for an exercise. I need to calculate this integral $$\displaystyle\int^x sin^4(t)dt$$

I don't really know where to start. Can you help me?

Thank you 🙂
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 06/11/2021 à 18:10:06

de TheLibrarian
**Modérateur**
Hi SDB2001 and welcome!

Before calculating the integral, you should rewrite $$sin^4(t)$$ in a more friendly form.

Do you know any formula to develop $$sin^4(t)$$?
Or at least $$sin^2(t)$$?
-------------------------- SDB2001

Membre
Posté le 06/11/2021 à 18:48:29

de SDB2001
Membre
I only know the formula for $$\displaystyle sin^2(\frac{\alpha}{2})$$.
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It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 06/11/2021 à 19:19:45

de TheLibrarian
**Modérateur**
Very good.
Try to develop $$sin^2(t)$$ using the formula you know. Then write $$(sin^2(t))^2$$ and see what you get.
-------------------------- SDB2001

Membre
Posté le 06/11/2021 à 20:33:07

de SDB2001
Membre
Ok, I developed and I get this

$$\displaystyle(sin^2(t))^2=\frac{1}{4}(1-cos(2t))^2$$

If I develop again, I now get this

$$\displaystyle(sin^2(t))^2=\frac{1}{4}(1-2cos(2t)+cos^2(2t))$$

But there is $$cos^2$$ now.
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 07/11/2021 à 05:42:12

de TheLibrarian
**Modérateur**
Don't you know a formula for $$cos^2(t)$$?
-------------------------- SDB2001

Membre
Posté le 07/11/2021 à 06:47:57

de SDB2001
Membre
Oh yes, I know $$\displaystyle cos^2(\frac{\alpha}{2})$$.

So if I use the formula in the calculation I get

$$\displaystyle sin^4(t)=\frac{1}{4}(1-2cos(2t)+1/2+cos(4t)/2) =\frac{1}{8}(3-4cos(2t)+cos(4t))$$

Is that correct?
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 07/11/2021 à 07:16:55

de TheLibrarian
**Modérateur**
Yes, absolutely. Now, you can easily calculate the initial integral. 😉
-------------------------- SDB2001

Membre
Posté le 07/11/2021 à 09:50:44

de SDB2001
Membre
Yes and I find that the integral is equal to

$$\displaystyle\frac{1}{8}(3x-2sin(2x)+\frac{1}{4}sin(4x))+constant$$

Is it the correct solution?
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 07/11/2021 à 10:00:27

de TheLibrarian
**Modérateur**
Very good. It is correct.
-------------------------- SDB2001

Membre
Posté le 07/11/2021 à 10:09:17

de SDB2001
Membre
Thank you very much for your help. 🙂
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 07/11/2021 à 10:23:16

de TheLibrarian
**Modérateur**
If you need more help, do not hesitate to ask again.
-------------------------- SDB2001

Membre
Posté le 07/11/2021 à 10:25:06

de SDB2001
Membre
Actually, I have other integrals to calculate. 😄
--------------------------
It's not rocket science but it looks like it
TheLibrarian

**Modérateur**
Posté le 07/11/2021 à 10:26:23

de TheLibrarian
**Modérateur**
No problem. Try and if you are stuck, come back and ask.
Good luck.
-------------------------- SDB2001

Membre
Posté le 07/11/2021 à 10:27:47

de SDB2001
Membre
Thank you again!
--------------------------
It's not rocket science but it looks like it
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