If-1-sin-x-4-sin-x-1-dx-A-1-tan-x-2-1-B-tan-1-f-x-C-then-

Question Number 53383 by gunawan last updated on 21/Jan/19

If \int \frac{1}{(\sin x + 4) (\sin x - 1)} d x

= A \frac{1}{\tan \frac{x}{2} - 1} + B \tan^{- 1} (f (x)) + C, then

Commented by maxmathsup by imad last updated on 21/Jan/19

l e t A = \int \frac{d x}{(s i n x + 4) (s i n x - 1)} \Rightarrow A =_{t a n (\frac{x}{2}) = t} \int \frac{1}{(\frac{2 t}{1 + t^{2}} + 4) (\frac{2 t}{1 + t^{2}} - 1)} \frac{2 d t}{1 + t^{2}}

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

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

l e t d e c o m p o s e F (t) = \frac{t^{2} + 1}{(2 t^{2} + t + 2) {(t - 1)}^{2}} \Rightarrow F (t) = \frac{a}{t - 1} + \frac{b}{{(t - 1)}^{2}} + \frac{c t + d}{2 t^{2} + t + 2}

b = {l i m}_{t \to 1} {(t - 1)}^{2} F (t) = \frac{2}{5}

{l i m}_{t \to + \infty} t F (t) = 0 = a + \frac{c}{2} \Rightarrow c = - 2 a \Rightarrow

F (t) = \frac{a}{t - 1} + \frac{2}{5 {(t - 1)}^{2}} + \frac{- 2 a t + d}{2 t^{2} + t + 2}

F (0) = \frac{1}{2} = - a + \frac{2}{5} + \frac{d}{2} \Rightarrow 1 = - 2 a + \frac{4}{5} + d \Rightarrow - 2 a + d = 1 - \frac{4}{5} = \frac{1}{5}

F (2) = \frac{5}{12} = a + \frac{2}{5} + \frac{- 4 a + d}{12} \Rightarrow 5 = 12 a + \frac{24}{5} - 4 a + d \Rightarrow 5 - \frac{24}{5} = 8 a + d \Rightarrow

8 a + d = \frac{1}{5} \Rightarrow d = \frac{1}{5} - 8 a \Rightarrow - 2 a + \frac{1}{5} - 8 a = \frac{1}{5} \Rightarrow a = 0 \Rightarrow d = \frac{1}{5} \Rightarrow

F (t) = \frac{2}{5 {(t - 1)}^{2}} + \frac{1}{5 (2 t^{2} + t + 2)} \Rightarrow A = - \frac{2}{5} \int \frac{d t}{{(t - 1)}^{2}} - \frac{1}{5} \int \frac{d t}{2 t^{2} + t + 2}

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

\int \frac{d t}{2 t^{2} + t + 2} = \frac{1}{2} \int \frac{d t}{t^{2} + 2 \frac{1}{4} t + \frac{1}{16} + 1 - \frac{1}{16}} = \frac{1}{2} \int \frac{d t}{{(t + \frac{1}{4})}^{2} + \frac{15}{16}}

=_{t + \frac{1}{4} = \frac{\sqrt{15}}{4} u} \frac{1}{2} \int \frac{1}{\frac{15}{16} (1 + u^{2})} \frac{\sqrt{15}}{4} d u = \frac{1}{2} . \frac{16}{15} . \frac{\sqrt{15}}{4} a r c t a n (u) + c_{2}

= \frac{2}{\sqrt{15}} a r c t a n (\frac{4 t + 1}{\sqrt{15}}) + c_{2} \Rightarrow A = \frac{2}{5} \frac{1}{t a n (\frac{x}{2}) - 1} - \frac{1}{5 \sqrt{15}} a r c t a n (\frac{4 t a n (\frac{x}{2}) + 1}{\sqrt{15}}) + C

\Rightarrow A = \frac{2}{5}, B = - \frac{1}{5 \sqrt{15}}, f (x) = \frac{4 a r c t a n (\frac{x}{2}) + 1}{\sqrt{15}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

\frac{1}{5} \int \frac{(s i n x + 4) - (s i n x - 1)}{(s i n x + 4) (s i n x - 1)} d x

\frac{1}{5} \int \frac{d x}{s i n x - 1} - \frac{1}{5} \int \frac{d x}{s i n x + 4}

\int \frac{d x}{a + s i n x}

\int \frac{d x}{a + \frac{2 t a n \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}

\int \frac{{s e c}^{2} \frac{x}{2} d x}{a + {a t a n}^{2} \frac{x}{2} + 2 t a n \frac{x}{2}}

k = t a n \frac{x}{2} 2 d k = {s e c}^{2} \frac{x}{2} d x

\int \frac{2 d k}{a + {a k}^{2} + 2 k}

\frac{2}{a} \int \frac{d k}{1 + k^{2} + 2 \frac{k}{a}}

\frac{2}{a} \int \frac{d k}{k^{2} + 2 \times k \times \frac{1}{a} + \frac{1}{a^{2}} + 1 - \frac{1}{a^{2}}}

\frac{2}{a} \int \frac{d k}{{(k + \frac{1}{a})}^{2} + {(\frac{\sqrt{a^{2} - 1}}{a})}^{2}}

\frac{2}{a} \times \frac{1}{(\frac{\sqrt{a^{2} - 1}}{a})} \times {t a n}^{- 1} (\frac{k + \frac{1}{a}}{\frac{\sqrt{a^{2} - 1}}{a}}) + c

= \frac{2}{\sqrt{a^{2} - 1}} {t a n}^{- 1} (\frac{a k + 1}{\sqrt{a^{2} - 1}}) + c

= \frac{2}{\sqrt{a^{2} - 1}} {t a n}^{- 1} (\frac{a t a n \frac{x}{2} + 1}{\sqrt{a^{2} - 1}}) + c

n o w \frac{1}{5} \int \frac{d x}{- 1 + s i n x} - \frac{1}{5} \int \frac{d x}{4 + s i n x}

= (- \frac{1}{5}) [\int \frac{d x}{1 - s i n x} + \int \frac{d x}{4 + s i n x}]

= (\frac{- 1}{5}) [\int \frac{1 + s i n x}{{c o s}^{2} x} d x + \frac{2}{\sqrt{16 - 1}} {t a n}^{- 1} (\frac{4 t a n \frac{x}{2} + 1}{\sqrt{4^{2} - 1}})]

= (\frac{- 1}{5}) [t a n x + s e c x + \frac{2}{\sqrt{15}} {t a n}^{- 1} (\frac{4 t a n \frac{x}{2} + 1}{\sqrt{15}})]

= (\frac{- 1}{5}) [\frac{\frac{2 t a n \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}} + 1}{\frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}} + d o]

= (\frac{- 1}{5}) [\frac{{(1 + t a n \frac{x}{2})}^{2}}{(1 + t a n \frac{x}{2}) (1 - t a n \frac{x}{2})} + d o]

= (\frac{- 1}{5}) (\frac{1 + t a n \frac{x}{2}}{1 - t a n \frac{x}{2}}) + (\frac{- 2}{5 \sqrt{15}}) ({t a n}^{- 1} (\frac{4 t a n \frac{x}{2} + 1}{\sqrt{15}})