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Question

11x3+|x|+3x2+4|x|+3dx is

A
4ππ20ln(sinα)dα
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B
4ππ20ln(cosα)dα
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C
2ππ20ln(sin2α)dα
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D
4ππ20(ln(sinα)+ln(cosα))dα
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Solution

The correct option is B 4ππ20ln(cosα)dα
11x3(|x|+1)(|x|+3)dx+11|x|+3(|x|+1)(|x|+3)dx
=0+2101x+1dx=2 ln2
Now π/20ln(sinα)dα=π/20ln(cosα)dα=π/20ln(sin2α)dα=π2ln2

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