How to integrate 1/(1+sinx)?

Question

I'm currently learning Trigonometric Functions and need help with solving two questions:

#int# #1/(1+sinx)# #dx# = ?

#int# #1/(1+cos2x)# #dx# = ?

Thanks!

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Answer 1 · 2017-10-02T07:40:00

#int1/(1+sinx)dx=tanx-secx# and #int1/(1+cos2x)dx=1/2tanx#

#int1/(1+sinx)dx#

= #int(1-sinx)/(1-sin^2x)dx#

= #int(1-sinx)/cos^2xdx#

= #int(sec^2x-tanxsecx)dx#

= #tanx-secx#

#int1/(1+cos2x)dx#

= #int1/(1+2cos^2x-1)dx#

= #int1/(2cos^2x)dx#

= #1/2intsec^2xdx#

= #1/2tanx#