Question
Mathematics Question on Integrals of Some Particular Functions
If ∫0π/2logcosxdx=2πlog(21) then ∫0π/2logsecxdx=
A
2πlog(1/2)
B
1−2πlog(1/2)
C
1+2πlog(1/2)
D
2πlog2
Answer
2πlog2
Explanation
Solution
Answer (d) 2πlog2
If ∫0π/2logcosxdx=2πlog(21) then ∫0π/2logsecxdx=
2πlog(1/2)
1−2πlog(1/2)
1+2πlog(1/2)
2πlog2
2πlog2
Answer (d) 2πlog2