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a
secxtanx+12logsecx+tanx+C
b
secxtanx−logsecx+tanx+C
c
12secxtanx+logsecx+tanx+C
d
12secxtanx+logsecx+tanx+C
answer is D.
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Detailed Solution
Let J=∫sec3xdxJ=∫secxsec2xdxUsing by partsJ=secx∫sec2dx−∫ddx(secx)∫sec2xdxdxJ=secxtanx−∫secxtanxtanxdxJ=secxtanx−∫secxsec2x−1dx2J=secxtanx+log|secx+tanx|J=12[secxtanx+log|secx+tanx|]