∫π20√cotx√cotx+√tanxdx=[MP PET 1990, 95; IIT 1983; MNR 1990]
A
π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
π3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cπ4 ∫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