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∫sin2x dx
Use the identity sin2x = ½ - ½(cos2x)
∫[½ - ½(cos2x)] dx = ∫½ dx - ∫½(cos2x) dx
Let's split it up into ∫½ dx and ∫½(cos2x) dx
∫½ dx = x/2 (we'll put the constant in at the end)
∫½(cos2x) dx (Use u substitution with u=2x and du = 2 dx)
∫cosu ¼du = ¼∫cosu du = ¼sinu + c = ¼sin2x (remember to resubstitute)
Subtract the two parts and add a constant
x/2 - ¼(sin2x) + c
This is also equivalent to: ½(x - sinxcosx) + c
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ā 12y ago
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What is the antiderivative of 2x?
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What is sine squared?
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