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Question

π20cos θsin3θ d θ=

A
2021
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B
821
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C
2021
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D
821
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Solution

The correct option is B 821
Let I=π20cos θsin3θ d θ
Put t=cos θdt=sin θ d θ,then
I=01t12(1t2)dt=10(t12t52)dt
I=[23t3227t72]10=821

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