Integral of sin(2*x)/(sin(x)^2+2) dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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The integral of a constant times a function is the constant times the integral of the function:
∫sin2(x)+22sin(x)cos(x)dx=2∫sin2(x)+2sin(x)cos(x)dx
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Let u=sin2(x)+2.
Then let du=2sin(x)cos(x)dx and substitute 2du:
∫2u1du
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The integral of a constant times a function is the constant times the integral of the function:
∫u1du=2∫u1du
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The integral of u1 is log(u).
So, the result is: 2log(u)
Now substitute u back in:
2log(sin2(x)+2)
So, the result is: log(sin2(x)+2)
Method #2
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Rewrite the integrand:
sin2(x)+2sin(2x)=sin2(x)+22sin(x)cos(x)
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The integral of a constant times a function is the constant times the integral of the function:
∫sin2(x)+22sin(x)cos(x)dx=2∫sin2(x)+2sin(x)cos(x)dx
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Let u=sin2(x)+2.
Then let du=2sin(x)cos(x)dx and substitute 2du:
∫2u1du
-
The integral of a constant times a function is the constant times the integral of the function:
∫u1du=2∫u1du
-
The integral of u1 is log(u).
So, the result is: 2log(u)
Now substitute u back in:
2log(sin2(x)+2)
So, the result is: log(sin2(x)+2)
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Now simplify:
log(sin2(x)+2)
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Add the constant of integration:
log(sin2(x)+2)+constant
The answer is:
log(sin2(x)+2)+constant
The answer (Indefinite)
[src]
/
|
| sin(2*x) / 2 \
| ----------- dx = C + log\sin (x) + 2/
| 2
| sin (x) + 2
|
/
∫sin2(x)+2sin(2x)dx=C+log(sin2(x)+2)
The graph
−log(3)+log(2)
=
−log(3)+log(2)
Use the examples entering the upper and lower limits of integration.