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Ans)

You can integrate tanθ

write tanθ=sinθ/cosθ

so now, integral of tanθ.dθ=integral of sinθ/cosθ.dθ→1

put cosθ=t→2

there fore -sinθ.dθ=dt→3

put 2 and 3 in 1

so,the question becomes integral of tanθ.dθ=integral of -dt/t→4

=-log t +c      →   5                  [reason:dx/x=logx]

put 2 in 5,

 =-log cosθ+C

where C is the integration constant

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