The value of ∫1/etanx t1+t2dt+∫1/ecotx dtt1+t2 is
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a
1/2
b
1
c
π/4
d
none of these
answer is B.
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Detailed Solution
1t1+t2=1t−t1+t2So the required integral is equal to∫1etanx tdx1+t2+∫1ecotx dtt−∫1ecotx tdt1+t2=12log1+t21etanx+logt1ecotx−12log1+t21ecotx=log|secx|+log|cotx|+1−log|cosecx|=1