integral of tanx/(sec x+cos x)?

Dear Hrithik Krrish,

Ans:- ∫tanx dx/(secx+cosx)

=∫sinx dx/(1+cos²x)

put cosx=z  So sinx dx = -dz

= -∫ dz/(1+z²)

= - tan¨¹z +c

= -tan¨¹(cosx) +c

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