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Q.

 1+2cotx(cosecx+cotx)dx0<x<π2 is equal to (where C is a constant of integration) 

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a

2 logsinx2+c

b

2 logcosx2+c

c

4 logsinx2+c

d

4 logcosx2+c

answer is B.

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Detailed Solution

  1+2cotx(cosecx+cotx)dx =1+2cot x cosecx+2cot2xdx =cosec2x+2cotx coxecx+cot2xdx =(cosec x+cot x)2dx =cosec x+cot xdx   0<x<π2 =1+cos xsin xdx =cotx2dx =2 log sinx2+C   

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