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`int(1-cosx)/(cosx(1+cosx))dx` |
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Answer» Correct Answer - `log|secx+tanx|-2tanx//2+C` `"Let " I=int(1-cosx)/(cosx(1+cosx))dx.` `"Let " cosx=y. " Then "` `(1-cosx)/(cosx(1+cosx))=(1-y)/(y(1+y))` `=(1)/(y)-(2)/(1+y)` `=(1)/(cosx)-(2)/(1+cosx)` `:. I=int(1-cosx)/(cosx(1+cosx))dx` `=int(1)/(cosx)dx-int(2)/(1+cosx)dx` `=int sec x dx-int(2)/(2cos^(2)x//2)dx` `=int secx-int sec^(2)x//2dx` `=log|secx+tanx|-2tanx//2+C` |
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