∫π20√cotx√cotx+√tanxdx=[MP PET 1990, 95; IIT 1983; MNR 1990]
A
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B
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C
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D
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Solution
The correct option is C ∫π20√cotx√cotx+√tanxdx⋯(i) =∫π20√cot(π2−x)√cot(π2−x)+√tan(π2−x)dx =∫π20√tanx√tanx+√cotxdx⋯(ii) Now adding (i) and (ii), we get 2I=∫π20√cotx+√tanx√tanx+√cotxdx=[x]π20⇒I=π4