tan-1-sin-d-

Question Number 64735 by mmkkmm000m last updated on 20/Jul/19

\int t a n θ / 1 \bar{+} s i n θ d θ

Commented by mr W last updated on 21/Jul/19

d e a r s i r :

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t h i s f o r u m, o t h e r w i s e w h y s h o u l d

p e o p l e h e l p o t h e r p e o p l e ?

Commented by mathmax by abdo last updated on 21/Jul/19

l e t I = \int \frac{t a n θ}{1 + s i n θ} d θ c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e

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

= - 4 \int \frac{4 t}{(t - 1) {(t + 1)}^{3}} d t l e t d e c o m p o s e F (t) = \frac{t}{(t - 1) {(t + 1)}^{3}}

F (t) = \frac{a}{t - 1} + \frac{b}{t + 1} + \frac{c}{{(t + 1)}^{2}} + \frac{d}{{(t + 1)}^{3}}

a = {l i m}_{x \to 1} (t - 1) F (t) = \frac{1}{8}

d = {l i m}_{t \to - 1} {(t + 1)}^{3} F (t) = \frac{1}{2} \Rightarrow F (t) = \frac{1}{8 (t - 1)} + \frac{b}{t + 1} + \frac{c}{{(t + 1)}^{2}} + \frac{1}{2 {(t + 1)}^{3}}

{l i m}_{t \to + \infty} t F (t) = 0 = a + b \Rightarrow b = - \frac{1}{8} \Rightarrow

F (t) = \frac{1}{8 (t - 1)} - \frac{1}{8 (t + 1)} + \frac{c}{{(t + 1)}^{2}} + \frac{1}{2 {(t + 1)}^{3}}

F (0) = 0 = - \frac{1}{8} - \frac{1}{8} + c + \frac{1}{2} = - \frac{1}{4} + \frac{1}{2} + c = \frac{1}{4} + c \Rightarrow c = - \frac{1}{4} \Rightarrow

F (t) = \frac{1}{8 (t - 1)} - \frac{1}{8 (t + 1)} - \frac{1}{4 {(t + 1)}^{2}} + \frac{1}{2 {(t + 1)}^{3}} \Rightarrow

\int F (t) d t = \frac{1}{8} l n ∣ t - 1 ∣ - \frac{1}{8} l n ∣ t + 1 ∣ + \frac{1}{4 (t + 1)} - \frac{1}{4 {(t + 1)}^{2}} + C

= \frac{1}{8} l n ∣ \frac{t - 1}{t + 1} ∣ + \frac{1}{4 (t + 1)} - \frac{1}{4 {(t + 1)}^{2}} + C \Rightarrow

I = - \frac{1}{2} l n ∣ \frac{t - 1}{t + 1} ∣ - \frac{1}{(t + 1)} + \frac{1}{{(t + 1)}^{2}} + C

= - \frac{1}{2} l n ∣ \frac{t a n (\frac{θ}{2}) - 1}{t a n (\frac{θ}{2}) + 1} ∣ - \frac{1}{1 + t a n (\frac{θ}{2})} + \frac{1}{{(1 + t a n (\frac{θ}{2}))}^{2}} + C .