Quack313
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#1
Report Thread starter 1 month ago
#1
2pi
∫ (6 - 8 sin θ cos θ) dθ = 12pi
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can some one help with this integral?
i also know that 8sinθcosθ can be written as 4sin(2θ)
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Shivesh777
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#2
Report 1 month ago
#2
You have done it yourself! Integral of 6 will give 12pi and area under the graph of 4sin(2 theta) from 0 to 2pi will be zero.
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