integral of (sec squarex+secx).dx to the limit 0 to pie/2

Asked by Sindhura K Reddy | 16th Aug, 2010, 09:10: PM

Expert Answer:

∫(sec2x + secx) dx =
∫sec2x dx + ∫secx dx =                   Using standard results,
tanx + ln(secx + tanx)             ... now put limits,
= tan π/2 + ln(sec π/2 + tan π/2) - tan 0 + ln(sec 0 + tan 0)
= ∞
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Team,
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Answered by  | 16th Aug, 2010, 09:57: PM

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