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Ans)
You can integrate tanθ
write tanθ=sinθ/cosθso now, integral of tanθ.dθ=integral of sinθ/cosθ.dθ→1put cosθ=t→2there fore -sinθ.dθ=dt→3put 2 and 3 in 1so,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+Cwhere C is the integration constant
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yes you can intergrate tan θ
tan q = sin q/cos q (1)
let,
y = cos q
so, dy/dq = -sin q
this implies, dq=dy/-sin q (2)
putting values in (1)
∫ 1/-y
= -log|y|
=-log|cos q|
so the ∫ tan q = -log|cos q|
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