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Question : 75 of 160
Marks:
+1,
-0
Solution:
Consider the integral,
I‌‌=‌cot9xdx ‌‌=‌‌dx Take
sin‌x=t then
cos‌x‌d‌x=dt So
I‌‌=1‌dt ‌‌=1‌dt ‌‌=1[t−9−4t−7+6t−5−4t−3+‌]dt ‌‌=[‌−‌+‌−‌+log‌t]‌1 Then
I‌‌=[(‌−‌+‌−‌+0)−(2−‌−6−4−‌‌log‌2)] ‌‌=‌+‌‌log‌2
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