Integral of -3ysinx+12ycosx dx
The solution
Detail solution
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫−3ysin(x)dx=−3y∫sin(x)dx
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The integral of sine is negative cosine:
∫sin(x)dx=−cos(x)
So, the result is: 3ycos(x)
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The integral of a constant times a function is the constant times the integral of the function:
∫12ycos(x)dx=12y∫cos(x)dx
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The integral of cosine is sine:
∫cos(x)dx=sin(x)
So, the result is: 12ysin(x)
The result is: 12ysin(x)+3ycos(x)
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Now simplify:
3y(4sin(x)+cos(x))
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Add the constant of integration:
3y(4sin(x)+cos(x))+constant
The answer is:
3y(4sin(x)+cos(x))+constant
The answer (Indefinite)
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| (-3*y*sin(x) + 12*y*cos(x)) dx = C + 3*y*cos(x) + 12*y*sin(x)
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∫(−3ysin(x)+12ycos(x))dx=C+12ysin(x)+3ycos(x)
Use the examples entering the upper and lower limits of integration.