Integral of (sqrt(1-2sinx))*cosx dx
The solution
Detail solution
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Let u=1−2sin(x).
Then let du=−2cos(x)dx and substitute −2du:
∫(−2u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫udu=−2∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=32u23
So, the result is: −3u23
Now substitute u back in:
−3(1−2sin(x))23
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Add the constant of integration:
−3(1−2sin(x))23+constant
The answer is:
−3(1−2sin(x))23+constant
The answer (Indefinite)
[src]
/
| 3/2
| ______________ (1 - 2*sin(x))
| \/ 1 - 2*sin(x) *cos(x) dx = C - -----------------
| 3
/
∫1−2sin(x)cos(x)dx=C−3(1−2sin(x))23
______________ ______________
/ /p\ / /p\ /p\
/ 1 - 2*sin|-| 2* / 1 - 2*sin|-| *sin|-|
1 \/ \6/ \/ \6/ \6/
- - ------------------ + ---------------------------
3 3 3
321−2sin(6p)sin(6p)−31−2sin(6p)+31
=
______________ ______________
/ /p\ / /p\ /p\
/ 1 - 2*sin|-| 2* / 1 - 2*sin|-| *sin|-|
1 \/ \6/ \/ \6/ \6/
- - ------------------ + ---------------------------
3 3 3
321−2sin(6p)sin(6p)−31−2sin(6p)+31
1/3 - sqrt(1 - 2*sin(p/6))/3 + 2*sqrt(1 - 2*sin(p/6))*sin(p/6)/3
Use the examples entering the upper and lower limits of integration.