E = ∫ 0 π / 2 1 − sin ( 2 0 1 6 x ) d x
Find E .
Bonus: Generalize for ∫ 0 π / 2 1 − sin ( 1 2 n x ) d x , where n is a positive integer , and prove it.
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Note another to find the integral of \rt(1-sinx) or is to multiply top and bottom by rt(1+sinx) which gives cosx/rt(1+sinx) which is an easy u sub to give 2rt(1+sinx)+C. Also works for rt(1-sinx)
sqrt(1 - sin 2016x) = abs(sin 1008x - cos 1008x), then treat as 504 separate integrations, each giving (2 x sqrt2)/1008). The first quarter cycle (0 to pi/4032) and the last three-quarter one can be treated as one whole cycle
Can you please explain how you got €=I/4
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I have changed the solution to explained it.
what if we make the integrand a perfect square an then simplify the two seperate integrals i.e (sin1008x-cos1008x)^2
It doesn' give the solutions even the integral calculator gives a different solution: CHECK THE FOLLOWING LINK BY PUTTING YOUR EXPRESSIONS...
https://www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B0%7D%5E%7B1.57%7D%5Csqrt%7B1-%5Csin%5Cleft(2016x%5Cright)%7Ddx
I'm trying to understand step 3 (dropping 2 k π due to periodicity and having k pop out of integral) and step 6, where we get cos out instead of sin. Any help?
Hello, first of all, I know I am wrong, but I am writing in order to somebody tells me where is my fail.
The primitive I achieve is 1008(Sqrt(1+sin(2016x)) and if we take the values x=0 and x=Pi/2, we obtain 0 as a result, which is absolutely false since the function is always positive. I know I am making a mistake, but I am not sure where...maybe in the process where cos(2016x) is involved?
Thank you for your help!!
Lemme just
E = ∫ 0 2 π 1 + sin ( 4 n x ) 1 − sin ( 4 n x ) 1 + sin ( 4 n x ) d x = ∫ 0 2 π 1 + sin ( 4 n x ) ∣ cos ( 4 n x ) ∣ d x sin ( 4 n x ) is periodic over 2 n π . = n ⋅ ( ∫ 0 8 n π 1 + sin ( 4 n x ) cos ( 4 n x ) d x − ∫ 8 n π 8 n 3 π 1 + sin ( 4 n x ) cos ( 4 n x ) d x + ∫ 8 n 3 π 2 n π 1 + sin ( 4 n x ) cos ( 4 n x ) d x ) t = sin ( 4 n x ) ⇔ d t = 4 n cos ( 4 n x ) d x = n ⋅ 4 n 1 ( ∫ 0 1 1 + t 1 d t − ∫ 1 − 1 1 + t 1 d t + ∫ − 1 0 1 + t 1 d t ) = 4 1 ( [ 2 1 + t ] 0 1 + [ 2 1 + t ] − 1 1 + [ 2 1 + t ] − 1 0 ) = [ 1 + t ] − 1 1 = 2 .
You have to justify the first step.
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What do you mean?
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It's not clear how my original integral is equal to his/her first line. S/he skipped some important steps here.
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@Hobart Pao – I don't see what's skipped.
I just multiplied 1 + sin ( 4 n x ) to the numerator and denominator,
and used the identity 1 + sin ( 4 n x ) 1 − sin ( 4 n x ) = 1 − sin 2 ( 4 n x ) = cos 2 ( 4 n x ) = ∣ cos ( 4 n x ) ∣ .
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@Boi (보이) – Nevermind, I'm stupid...sorry.
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@Hobart Pao – It's okay!
And nah you're not, everyone makes mistakes ^^;
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E = ∫ 0 2 π 1 − sin ( 4 k x ) d x = 4 k 1 ∫ 0 2 k π 1 − sin θ d θ = 4 k k ∫ 0 2 π 1 − sin θ d θ = 4 1 ( ∫ 0 π 1 − sin θ d θ + ∫ π 2 π 1 − sin θ d θ ) = 4 1 ( 2 ∫ 0 2 π 1 − sin θ d θ + 2 ∫ π 2 3 π 1 − sin θ d θ ) = 2 1 ( ∫ 0 2 π 1 − sin ( 2 π − θ ) d θ + ∫ π 2 3 π 1 − sin ( 2 π − θ ) d θ ) = 2 1 ( ∫ 0 2 π 1 − cos θ d θ + ∫ π 2 3 π 1 − cos θ d θ ) = 2 1 ( ∫ 0 2 π 2 sin 2 θ d θ + ∫ π 2 3 π 2 sin 2 θ d θ ) = 2 ( cos 2 θ ∣ ∣ ∣ ∣ 2 π 0 + cos 2 θ ∣ ∣ ∣ ∣ 2 3 π π ) = 2 ( 1 − 2 1 + 0 + 2 1 ) = 2 Let θ = 4 k x ⟹ d θ = 4 k d x Since the integrand has a period of 2 π See note. Since sin θ is symmetrical about 2 π and 2 3 π Using identity ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x Since cos 2 u = 1 − 2 sin 2 u
We note that the result applies to all multiples of 4, which 2016 is one. Therefore, ⟹ ⌊ 1 0 0 0 E ⌋ = 1 4 1 4
Note:
The figure is a plot of f ( x ) = 1 − sin θ , we note that the area under the curve from 0 to 2 k π or k cycles is the same as the area under the curve from 0 to 2 π or 1 cycle multiplies by k .
⟹ ∫ 0 2 k π f ( x ) d x = k ∫ 0 2 π f ( x ) d x