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Question

# The value of integral ∫π3π6sinx−xcosxx(x+sinx)dx is

A
loge2(π+3)(2π+33)
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B
loge(π+3)2(2π+33)
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C
loge(2π+33)2(π+3)
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D
loge2(2π+33)(π+3)
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Solution

## The correct option is B loge⎧⎪⎨⎪⎩2(π+3)(2π+3√3)⎫⎪⎬⎪⎭Let I=∫π3π6sinx−xcosxx(x+sinx)dx⇒I=∫π3π6(x+sinx)−x(1+cosx)x(x+sinx)dx⇒I=∫π3π6(1x−1+cosxx+sinx)dx⇒I=[logx]π/3π/6−∫π3π61+cosxx+sinxdxPut t=x+sinx; dt=(1+cosx)dx in 2nd term⇒I=(logπ3−logπ6)−∫(π3+√32)(π6+12)dtt⇒I=log2−[logt](π3+√32)(π6+12)⇒I=log2−[log(π3+√32)−log(π6+12)]⇒I=log2−log(2π+3√3π+3)(∵logm−logn=logmn)I=log(2(π+3)2π+3√3)=log(2π+62π+3√3)

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