Question

integral of tanx/logsecx​

Answer 1

Answer:

ANSWER

∫e

x

(tanx+logsecx)dx

=∫e

x

tanxdx+∫e

x

logsecxdx

=∫e

x

tanxdx+log(secx)e

x

−∫

dx

d

(logsecx).∫e

x

dx

=∫e

x

tanxdx+log(sec)xe

x

−∫e

x

tanxdx

=e

x

logsecx+c

Step-by-step explanation:

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