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Question

π20cot xcot x+tan xdx= [MP PET 1990, 95; IIT 1983; MNR 1990]


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Solution

The correct option is C
π20cot xcot x+tan xdx (i)
=π20cot(π2x)cot(π2x)+tan(π2x)dx
=π20tan xtan x+cot xdx (ii)
Now adding (i) and (ii), we get
2I=π20cot x+tan xtan x+cot x dx=[x]π20I=π4

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