Integral of -sin2x dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫(−sin(2x))dx=−∫sin(2x)dx
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There are multiple ways to do this integral.
Method #1
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Let u=2x.
Then let du=2dx and substitute 2du:
∫2sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫sin(u)du=2∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −2cos(u)
Now substitute u back in:
−2cos(2x)
Method #2
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The integral of a constant times a function is the constant times the integral of the function:
∫2sin(x)cos(x)dx=2∫sin(x)cos(x)dx
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There are multiple ways to do this integral.
Method #1
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Let u=cos(x).
Then let du=−sin(x)dx and substitute −du:
∫(−u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫udu=−∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −2u2
Now substitute u back in:
−2cos2(x)
Method #2
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Let u=sin(x).
Then let du=cos(x)dx and substitute du:
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
Now substitute u back in:
2sin2(x)
So, the result is: −cos2(x)
So, the result is: 2cos(2x)
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Add the constant of integration:
2cos(2x)+constant
The answer is:
2cos(2x)+constant
The answer (Indefinite)
[src]
/
| cos(2*x)
| -sin(2*x) dx = C + --------
| 2
/
∫(−sin(2x))dx=C+2cos(2x)
The graph
Use the examples entering the upper and lower limits of integration.