tan-x-1-Cos-x-dx-Need-help-

Question Number 25335 by mubeen897@hotmail.com last updated on 08/Dec/17

\int \frac{t a n (x)}{(1 + C o s (x))} d x = ?

N e e d h e l p ? ?

Answered by ajfour last updated on 08/Dec/17

I = \int \frac{\sin x}{\cos x (1 + \cos x)} d x

l e t \cos x = t \Rightarrow - \sin x d x = d t

\Rightarrow I = - \int \frac{d t}{t (1 + t)} = \int \frac{d t}{1 + t} - \int \frac{d t}{t}

= \ln ∣ \frac{1 + t}{t} ∣ + c

\Rightarrow I = \ln ∣ \frac{1 + \cos x}{\cos x} ∣ + c .

Answered by $@ty@m last updated on 08/Dec/17

= \int \frac{\sin x}{\cos x (1 + \cos x)} d x

= \int \frac{- d t}{t (1 + t)}

= \int \frac{- d t}{t^{2} + t + \frac{1}{4} - \frac{1}{4}}

= \int \frac{- d t}{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}

= \int \frac{d t}{{(\frac{1}{2})}^{2} - {(t + \frac{1}{2})}^{2}}

= \frac{1}{2 \times \frac{1}{2}} \ln ∣ \frac{\frac{1}{2} + t + \frac{1}{2}}{\frac{1}{2} - t - \frac{1}{2}} ∣

= \ln ∣ \frac{t + 1}{t} ∣ + C

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