calculate-0-pi-dx-cosx-2sinx-2-

Question Number 85801 by mathmax by abdo last updated on 24/Mar/20

c a l c u l a t e \int_{0}^{π} \frac{d x}{{(c o s x + 2 s i n x)}^{2}}

Commented by john santu last updated on 25/Mar/20

{(\cos x + 2 \sin x)}^{2} = 5 \cos^{2} (x - \tan^{- 1} (2))

\int \frac{d x}{5 \cos^{2} (x - \tan^{- 1} (2))} =

\frac{1}{5} \tan (x - \tan^{- 1} (2)) =

\frac{1}{5} (\frac{\tan x - 2}{1 + 2 \tan x}) + c

s o \Rightarrow \underset{0}{\int^{π}} \frac{d x}{{(\cos x + 2 \sin x)}^{2}} =

Missing \left or extra \right

\frac{1}{5} [- 2 - (- 2)] = 0

Commented by mathmax by abdo last updated on 26/Mar/20

e t A = \int_{0}^{π} \frac{d x}{{(c o s x + 2 s i n x)}^{2}} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

A = \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) {\frac{1 - t^{2}}{1 + t^{2}} + \frac{4 t}{1 + t^{2}}}^{2}} = \int_{0}^{\infty} \frac{2 (1 + t^{2})}{{(- t^{2} + 4 t + 1)}^{2}} d t

= 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} - 4 t - 1)}^{2}} d t l e t d e c o m p o s e F (t) = \frac{t^{2} + 1}{{(t^{2} - 4 t - 1)}^{2}}

t^{2} - 4 t - 1 = 0 \to Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5}

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

b = \frac{t_{1}^{2} + 1}{{(t_{1} - t_{2})}^{2}} = \frac{4 + 4 \sqrt{5} + 5 + 1}{{(2 \sqrt{5})}^{2}} = \frac{10 + 4 \sqrt{5}}{20} = \frac{5 + 2 \sqrt{5}}{10}

d = \frac{t_{2}^{2} + 1}{{(t_{2} - t_{1})}^{2}} = \frac{4 - 4 \sqrt{5} + 5 + 1}{20} = \frac{10 - 4 \sqrt{5}}{20} = \frac{5 - 2 \sqrt{5}}{10}

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

F (0) = \frac{1}{{(t_{1} t_{2})}^{2}} = 1 = - \frac{a}{t_{1}} + \frac{b}{t_{1}^{2}} + \frac{a}{t_{2}} + \frac{d}{t_{2}^{2}} \Rightarrow

(\frac{1}{t_{2}} - \frac{1}{t_{1}}) a = 1 - \frac{b}{t_{1}^{2}} - \frac{d}{t_{2}^{2}} \Rightarrow (\frac{2 \sqrt{5}}{- 1}) a = 1 - \frac{b}{4 + 4 \sqrt{5} + 5} - \frac{d}{4 - 4 \sqrt{5} + 5}

= 1 - \frac{b}{9 + 4 \sqrt{5}} - \frac{d}{9 - 4 \sqrt{5}} \Rightarrow a = - \frac{1}{2 \sqrt{5}} {1 - \frac{b}{9 + 4 \sqrt{5}} - \frac{d}{9 - 4 \sqrt{5}}} \Rightarrow

A = 2 a \int_{0}^{\infty} \frac{d t}{t - t_{1}} + 2 b \int_{0}^{\infty} \frac{d t}{{(t - t_{1})}^{2}} + 2 c \int_{0}^{\infty} \frac{d t}{t - t_{2}} + 2 d \int_{0}^{\infty} \frac{d t}{{(t - t_{2})}^{2}}

= {[2 a l n ∣ t - t_{1} ∣ + 2 c l n ∣ t - t_{2} ∣]}_{0}^{+ \infty} + {[\frac{- 2 b}{t - t_{1}} - \frac{2 d}{t - t_{2}}]}_{0}^{+ \infty}

r e s t t o f i n i s h t h e c a l c u l u s \dots .

Commented by mathmax by abdo last updated on 26/Mar/20

l e t t r y a n o t h e r w a y w e h a v e

A = 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t - t_{1})}^{2} {(t - t_{2})}^{2}} w i t h t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5}

A = 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(\frac{t - t_{1}}{t - t_{2}})}^{2} {(t - t_{2})}^{4}} d t w e d o t h e c h a n g e m e n t \frac{t - t_{1}}{t - t_{2}} = u \Rightarrow

t - t_{1} = u t - {u t}_{2} \Rightarrow (1 - u) t = t_{1} - {u t}_{2} \Rightarrow t = \frac{t_{1} - {u t}_{2}}{1 - u} \Rightarrow

d t = \frac{- t_{2} (1 - u) + (t_{1} - {u t}_{2})}{{(1 - u)}^{2}} = \frac{- t_{2} + t_{2} u + t_{1} - {u t}_{2}}{{(1 - u)}^{2}} = \frac{2 \sqrt{5}}{{(1 - u)}^{2}}

t - t_{2} = \frac{t_{1} - {u t}_{2}}{1 - u} - t_{2} = \frac{t_{1} - {u t}_{2} - t_{2} + {u t}_{2}}{1 - u} = \frac{2 \sqrt{5}}{1 - u} \Rightarrow

A = 2 \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{{(\frac{t_{1} - {u t}_{2}}{1 - u})}^{2} + 1}{u^{2} \times \frac{{(2 \sqrt{5})}^{4}}{{(1 - u)}^{4}}} \times \frac{2 \sqrt{5}}{{(1 - u)}^{2}} d u

A = \frac{2}{{(2 \sqrt{5})}^{3}} \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{{(t_{1} - {u t}_{2})}^{2} + {(1 - u)}^{2}}{u^{2}} d u

A = \frac{1}{20 \sqrt{5}} \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{t_{2}^{2} u^{2} + 2 u + t_{1}^{2} + u^{2} - 2 u + 1}{u^{2}} d u (t_{1} t_{2} = - 1)

20 \sqrt{5} A = \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{(1 + t_{2}^{2}) u^{2} + t_{1}^{2}}{u^{2}} d u

= (1 + t_{2}^{2}) (1 - \frac{t_{1}}{t_{2}}) - t_{1}^{2} {[\frac{1}{u}]}_{\frac{t_{1}}{t_{2}}}^{1} = (1 + t_{2}^{2}) \frac{t_{2} - t_{1}}{t_{2}} - t_{1}^{2} (1 - \frac{t_{2}}{t_{1}}) \Rightarrow

A = \frac{1}{20 \sqrt{5}} {(1 + {(2 - \sqrt{5})}^{2}) (\frac{- 2 \sqrt{5}}{2 - \sqrt{5}}) - {(2 + \sqrt{5})}^{2} (\frac{2 \sqrt{5}}{2 + \sqrt{5}})}