Q.

∫0π4(πx−4x2)ln(1+tanx)dx=

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a

π296ln2

b

π3192ln2

c

π296

d

π3192

answer is B.

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Detailed Solution

I=4∫0π4x(π4−x)ln(1+tanx)dx,x→π4−x I=4∫0π4(π4−x)xln(1+tan(π4−x))dx I=4∫0π4x(π4−x)xln(21+tanx)dx 2I=4ln2∫0π4x(π4−x)dx I=2ln2[(π4)3⋅12−(π4)3⋅13] =π3192ln2
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∫0π4(πx−4x2)ln(1+tanx)dx=