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Question : 154 of 160
Marks:
+1,
-0
Solution:
Let
I=∫dx =∫dx−∫dx =‌∫dx−∫dx =‌∫xsec2()dx−∫tan()‌d‌x ={x.2‌tan()−∫2‌tan()‌d‌x} −∫tan()‌d‌x =x‌tan‌−∫tan‌‌d‌x −∫tan()+C =x‌tan‌−4‌log|sec‌|+C but given,
∫dx =x‌tan‌+p‌log|sec‌|+C On comparing, we get
p=−4
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