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Question

lf I1=tanx1/et1+t2dt and I2iscotx1/edtt(1+t2) then the value of I1+I2 is

A
12
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B
1
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C
e2
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D
12(e+1e)
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Solution

The correct option is D 1
I1=12tanx1e2t1+t2dt=12[log(1+t2)]tanx1e=12[log(1+tan2x)log(1+1e2)]=12[log(sec2x)log(e2+1)log(e2)]=logsecxlog(e2+1)4I2=cotx1edtt(1+t2)=cotx1edtt12cotx1e2t1+t2dt=[log]cotx1e12[log(1+e2)]cotx1e=log(cotx)log1e12[logcosecxlog(1+1e2)]=log(cotx)+1logcosecx+14log(e2+14)=logsinx+logcosx+1+logsinx+14log(e2+14)[logcosecx=log(cosecx)1=logsinx][logcotx=log(cosecx)+1+14log(e2+14)]I1+I2=1+2=logcosx+1+log(secx)=logcosx+1logsecx=1

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