Question

integration of 1-tanx upon 1+tanx

Answer 1

hello friend..!!

$\int\limits (1 - tanx ) / ( 1 + tanx) \, dx$ 

implies,

⇒ $\int\limits (cosx + sinx) / (cosx - sinx) \, dx$

⇒ now by putting cosx - sinx = t 

⇒( - sinx - cosx )dx = dt 

⇒ ( sinx + cosx ) = -dt

⇒ I = -$\int\limits dt/t$

⇒ -log| t | + c 

⇒-log | cosx - sinx | + c .

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hope it is useful....!!