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Course: Integral Calculus>Unit 1

Lesson 15: Integrating using trigonometric identities

Integral of sin^4(x)

It is a bit involved, but we can use u-substitution to find the integral of sin^4(x).

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• Why does sin^2(x) = (1/2)(1-cos(2x) not (1-cos^2(x)) ?
• It does, however converting from one trig function that is squared to another that is squared doesn't get you any further in solving the problem. But converting a squared trig function to one that isn't squared, such as in the video, well, sin²x gets you 1/2 - cos(2x)/2, and that you can integrate directly.
• At around , why does (1-cos(2x))² turn into 1-2cos(2x)+cos²(2x) ? Where does the 2cos(2x) come from ?
• Sal was simply expanding the expression (1/2 (1-cos2x))^2. 1/2 squared is 1/4 and (1-cos2x) squared is (1-cos2x) times (1-cos2x). If you recall from Algebra that (a-b)(a-b) = a^2 -2ab + b^2, then let a = 1 and let b = cos2x and multiply it out. a^2 = (1)(1) = 1, -2ab = -2(1)(cos2x) = -2cos2x. b^2 = (cos2x)(cos2x) = cos^2 2x. Hope this helps. Good luck.
• Do i have to memorize the double angle identity formula? I searched the trig identities formula. But there are so many.. :(
• (In case anyone wanted) A quick derivation for the trig identity {sin^2x=1/2(1-cos2x)} :
2. Rearrange both: sin^2x=1-cos^2x and cos^2x=cos2x+sin^2x
3. Substitute cos2x+sin^2x into sin^2x=1-cos^2x for cos^2x
4. Expand: sin^2x=1-cos2x-sin^2x
5. Add sin^2x to both sides, giving 2sin^2x=1-cos2x
6. Divide both sides by 2, leaving sin^2x= 1/2(1-cos2x)
• At , how do you get (1/8) *4cos4x ??
• Where does the Trig identity at come from?
• at i dont understand why the 2 disappeared after you integrate the 2 cos2x which ibecamesin 2x..same as 4cos4x which became sin4x.
• Is
sin^2(x)=(1-cos(2x))/2
&
cos^2(x)=(1+cos(4x)/2
another two trig function properties? I've never met them before and wonder if anyone can provide me the link to these properties, either articles or vids.
• Please, can anyone explain how (sinAx)(sinBx) = 1/2(cos(Ax-Bx)-cos(Ax+Bx)) Thanks
• Difference Formula:
cos(Ax - Bx) = cos(Ax)cos(Bx) + sin(Ax)sin(Bx) *
Sum Formula:
cos(Ax + Bx) = cos(Ax)cos(Bx) - sin(Ax)sin(Bx)
<=> - cos(Ax + Bx) = - cos(Ax)cos(Bx) + sin(Ax)sin(Bx) **
Add equations * and **. We get
cos(Ax - Bx) - cos(Ax + Bx) = 2sin(Ax)sin(Bx)
<=> sin(Ax)sin(Bx) = ½(cos(Ax - Bx) - cos(Ax + Bx))
• In my trigonometry book, there is the following identity:
sin P + sin Q
= 2sin((P + Q)/2) cos((P – Q)/2)
and also one for sin P – sin Q and cos P ± cos Q. It's mentioned that this identity can be helpful in integration, but I can't see how this identity can be helpful, especially in integration. Can someone please tell me a case where this could be helpful? Thanks in advance.