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Calculation of the integral of sine to power 4
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Author Message
SDB2001

Member
Posted on 06/11/2021 at 17:45:53

by SDB2001
Member
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

**Moderator**
Posted on 06/11/2021 at 18:10:06

by TheLibrarian
**Moderator**
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

Member
Posted on 06/11/2021 at 18:48:29

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

**Moderator**
Posted on 06/11/2021 at 19:19:45

by TheLibrarian
**Moderator**
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

Member
Posted on 06/11/2021 at 20:33:07

by SDB2001
Member
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

**Moderator**
Posted on 07/11/2021 at 05:42:12

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

Member
Posted on 07/11/2021 at 06:47:57

by SDB2001
Member
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

**Moderator**
Posted on 07/11/2021 at 07:16:55

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

Member
Posted on 07/11/2021 at 09:50:44

by SDB2001
Member
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

**Moderator**
Posted on 07/11/2021 at 10:00:27

by TheLibrarian
**Moderator**
Very good. It is correct.
-------------------------- SDB2001

Member
Posted on 07/11/2021 at 10:09:17

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

**Moderator**
Posted on 07/11/2021 at 10:23:16

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

Member
Posted on 07/11/2021 at 10:25:06

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

**Moderator**
Posted on 07/11/2021 at 10:26:23

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

Member
Posted on 07/11/2021 at 10:27:47

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