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Calculation of the integral of sine to power 4

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Author | Message |
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SDB2001
Member |
Posted on 06/11/2021 at 17:45:53
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
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
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
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
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
Don't you know a formula for \(cos^2(t)\)?
-------------------------- |

SDB2001
Member |
Posted on 07/11/2021 at 06:47:57
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
Yes, absolutely. Now, you can easily calculate the initial integral. 😉
-------------------------- |

SDB2001
Member |
Posted on 07/11/2021 at 09:50:44
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
Very good. It is correct.
-------------------------- |

SDB2001
Member |
Posted on 07/11/2021 at 10:09:17
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
If you need more help, do not hesitate to ask again.
-------------------------- |

SDB2001
Member |
Posted on 07/11/2021 at 10:25:06
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
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
Thank you again!
-------------------------- It's not rocket science but it looks like it |

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