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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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โˆ™ 2012-01-14 20:21:33
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