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Question : 73 of 160
Marks:
+1,
-0
Solution:
Solve as follow,
I=∫=∫| sec4x |
| (2‌tan‌x+sec2x)4 |
dx=∫sec2xdxLet
tan‌x=tSo,
sec2xdx=dtSo,
I=dt=dt=∫−2‌∫+2‌∫=−(1+tan‌x)−5+(1+tan‌x)−6 −(1+tan‌x)−7+kSo,
A=−,B= and
C=−⇒
A+B+C=−
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