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Question : 51 of 160
Marks:
+1,
-0
Solution:
∫(sec2−tan‌)dx=‌∫(xsec2)dx=∫tan‌‌d‌x={x(2‌tan‌)−1‌∫2‌tan‌‌d‌x} −∫tan‌‌d‌x =x‌tan‌−2‌∫tan‌‌d‌x=x‌tan‌−2+c=x‌tan‌−4‌log|sec‌|+c ‌‌p=−4
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