Integral of (sinx+cosx)/(3+sin2x) dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Rewrite the integrand:
sin(2x)+3sin(x)+cos(x)=sin(2x)+3sin(x)+sin(2x)+3cos(x)
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Integrate term-by-term:
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Don't know the steps in finding this integral.
But the integral is
∫sin(2x)+3sin(x)dx
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Don't know the steps in finding this integral.
But the integral is
∫sin(2x)+3cos(x)dx
The result is: ∫sin(2x)+3sin(x)dx+∫sin(2x)+3cos(x)dx
Method #2
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Rewrite the integrand:
sin(2x)+3sin(x)+cos(x)=2sin(x)cos(x)+3sin(x)+2sin(x)cos(x)+3cos(x)
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Integrate term-by-term:
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Don't know the steps in finding this integral.
But the integral is
42(atan(22tan(2x)−22)+π⌊π2x−2π⌋)−42(atan(232tan(2x)+22)+π⌊π2x−2π⌋)−8log(tan2(2x)−2tan(2x)+3)+8log(9tan2(2x)+6tan(2x)+3)
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Don't know the steps in finding this integral.
But the integral is
−42(atan(22tan(2x)−22)+π⌊π2x−2π⌋)+42(atan(232tan(2x)+22)+π⌊π2x−2π⌋)−8log(tan2(2x)−2tan(2x)+3)+8log(9tan2(2x)+6tan(2x)+3)
The result is: −4log(tan2(2x)−2tan(2x)+3)+4log(9tan2(2x)+6tan(2x)+3)
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Add the constant of integration:
∫sin(2x)+3sin(x)dx+∫sin(2x)+3cos(x)dx+constant
The answer is:
∫sin(2x)+3sin(x)dx+∫sin(2x)+3cos(x)dx+constant
The answer (Indefinite)
[src]
/ / /
| | |
| sin(x) + cos(x) | cos(x) | sin(x)
| --------------- dx = C + | ------------ dx + | ------------ dx
| 3 + sin(2*x) | 3 + sin(2*x) | 3 + sin(2*x)
| | |
/ / /
∫sin(2x)+3sinx+cosxdx
1
/
|
| cos(x) + sin(x)
| --------------- dx
| 3 + sin(2*x)
|
/
0
∫01sin(2x)+3sinx+cosxdx
=
1
/
|
| cos(x) + sin(x)
| --------------- dx
| 3 + sin(2*x)
|
/
0
0∫1sin(2x)+3sin(x)+cos(x)dx
Use the examples entering the upper and lower limits of integration.