∫secx(1+tanx)dx(e−x+secx)=f(x)+c. If f(0)=ln(2), then f(π4) is
A
ln(1+√2eπ/4)
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B
ln√2
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C
ln(2√2)
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D
ln(eπ/4√2+1)
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Solution
The correct option is Aln(1+√2eπ/4) ∫secx(1+tanx)dx(e−x+secx) =∫exsecx(1+tanx)dx(1+exsecx) Let1+exsecx=t ⇒exsecx(1+tanx)dx=dt ∫dtt=lnt+c=ln(1+exsecx)+c If f(0)=ln(2) then c=0 So f(x)=ln(1+exsecx) f(π4)=ln(1+√2eπ/4)