∫ ((1−sin 2x)/((1+sin 2x)^2 )) dx =?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 168188 by cortano1 last updated on 05/Apr/22

$\int \frac{1 - \sin 2 x}{{(1 + \sin 2 x)}^{2}} d x = ?$
Commented by Florian last updated on 09/Apr/22

$? = \frac{1}{2 \sqrt{2}} a r c t a n (\sqrt{2} t a n (2 x)) + \frac{1}{4 \sqrt{2}} (l n ∣ \frac{1}{\sqrt{2}} c o s (2 x) + 1 ∣ l n ∣ \frac{1}{\sqrt{2}} c o s (2 x) - 1 ∣) + c, c \in R$
	Answered by peter frank last updated on 05/Apr/22

	$1 - \sin 2 x = {(\cos x - \sin x)}^{2}$ ${(1 + \sin 2 x)}^{2} = {(\cos x + \sin x)}^{4}$ $\int \frac{{(\cos x - \sin x)}^{2}}{{(\cos x + \sin x)}^{4}} dx$ $\int \frac{{(1 - \tan x)}^{2}}{{(1 + \tan x)}^{4}} dx . \sec^{2} x$ $t = \tan x dx = \frac{1}{\sec^{2} x} dt$ $\int \frac{{(1 - t)}^{2}}{{(1 + t)}^{4}} dt$ $. . . . . . .$
	Answered by Mathspace last updated on 05/Apr/22
	$I=_(2x=t) (1/2) ∫ ((1−sint)/((1+sint)^2 ))dt ⇒ f(λ)=∫ ((1−sint)/(λ+sint))dt ⇒ f^′ (λ)=−∫ ((1−sint)/((λ+sint)^2 ))dt ⇒ ∫ ((1−sint)/((1+sint)^2 ))=−f^′ (1) f(λ)=_(tan((t/2))=x) ∫ ((1−((2x)/(1+x^2 )))/(λ+((2x)/(1+x^2 ))))((2dx)/(1+x^2 )) =2∫ ((1+x^2 −2x)/((1+x^2 ){λ+λx^2 +2x}))dx we decompose F(x)=((x^2 −2x+1)/((x^2 +1)(λx^2 +2x+λ))) F(x)=((ax+b)/(x^2 +1))+((cx+d)/(λx^2 +2x+λ)) ⇒(ax+b)(λx^2 +2x+λ)+(cx+d)(x^2 +1) =x^2 −2x+1 ⇒ λax^3 +2ax^2 +λax +λbx^2 +2bx+λb cx^3 +cx+dx^2 +d=x^2 −2x+1 ⇒ (λa+c)x^3 +(2a+λb+d)x^2 +(λa+2b+c)x +λb+d=x^2 −2x+1 ⇒ λa+c=o ,2a+λb+d=1 ,λa+2b+c=−2 λb+d=1....be continued....$
	$I =_{2 x = t} \frac{1}{2} \int \frac{1 - s i n t}{{(1 + s i n t)}^{2}} d t \Rightarrow$ $f (λ) = \int \frac{1 - s i n t}{λ + s i n t} d t \Rightarrow$ $f^{^{'}} (λ) = - \int \frac{1 - s i n t}{{(λ + s i n t)}^{2}} d t \Rightarrow$ $\int \frac{1 - s i n t}{{(1 + s i n t)}^{2}} = - f^{^{'}} (1)$ $f (λ) =_{t a n (\frac{t}{2}) = x} \int \frac{1 - \frac{2 x}{1 + x^{2}}}{λ + \frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}$ $= 2 \int \frac{1 + x^{2} - 2 x}{(1 + x^{2}) {λ + λ x^{2} + 2 x}} d x$ $w e d e c o m p o s e F (x) = \frac{x^{2} - 2 x + 1}{(x^{2} + 1) (λ x^{2} + 2 x + λ)}$ $F (x) = \frac{a x + b}{x^{2} + 1} + \frac{c x + d}{λ x^{2} + 2 x + λ}$ $\Rightarrow (a x + b) (λ x^{2} + 2 x + λ) + (c x + d) (x^{2} + 1)$ $= x^{2} - 2 x + 1 \Rightarrow$ $λ {a x}^{3} + 2 {a x}^{2} + λ a x + λ {b x}^{2} + 2 b x + λ b$ ${c x}^{3} + c x + {d x}^{2} + d = x^{2} - 2 x + 1 \Rightarrow$ $(λ a + c) x^{3} + (2 a + λ b + d) x^{2} + (λ a + 2 b + c) x$ $+ λ b + d = x^{2} - 2 x + 1 \Rightarrow$ $λ a + c = o, 2 a + λ b + d = 1, λ a + 2 b + c = - 2$ $λ b + d = 1 . . . . b e c o n t i n u e d . . . .$
	Answered by MJS_new last updated on 06/Apr/22

	$\int \frac{1 - \sin 2 x}{{(1 + \sin 2 x)}^{2}} d x = \int (\frac{2}{{(1 + \sin 2 x)}^{2}} - \frac{1}{1 + \sin 2 x}) d x =$ $[t = \tan x \to d x = \frac{d t}{t^{2} + 1}]$ $= \int (\frac{2 (t^{2} + 1)}{{(t + 1)}^{4}} - \frac{1}{{(t + 1)}^{2}}) d t =$ $= \int (\frac{4}{{(t + 1)}^{4}} - \frac{4}{{(t + 1)}^{3}} + \frac{1}{{(t + 1)}^{2}}) d t =$ $= - \frac{4}{3 {(t + 1)}^{3}} + \frac{2}{{(t + 1)}^{2}} - \frac{1}{t + 1} =$ $= - \frac{3 t^{2} + 1}{3 {(t + 1)}^{3}} =$ $= - \frac{1 + {3 \tan}^{2} x}{3 {(1 + \tan x)}^{3}} + C$ $[= - \frac{\cos^{3} 2 x}{6 {(1 + \sin 2 x)}^{3}} + C]$
	Commented by peter frank last updated on 06/Apr/22

	$thank you$
	Answered by greogoury55 last updated on 06/Apr/22
	$T=∫ ((1−sin 2x)/((1+sin 2x)^2 )) dx [ tan x=t → { ((dx=(dt/(1+t^2 )))),((sin 2x=((2tan x)/(1+tan^2 x))=((2t)/(1+t^2 )))) :}] T=∫ ((1+t^2 −2t)/((1+t^2 +2t)^2 )) dt =∫ (((t−1)^2 )/((t+1)^4 )) dt = ∫ ((1/((t+1)^2 )) −(4/((t+1)^3 )) +(4/((t+1)^4 )))dt = −(1/(t+1)) +(2/((t+1)^2 ))−(4/(3(t+1)^3 )) + c = −(1/(1+tan x)) +(2/((1+tan x)^2 )) −(4/((1+tan x)^3 )) + c$
	$T = \int \frac{1 - \sin 2 x}{{(1 + \sin 2 x)}^{2}} d x$ $[\tan x = t \to {\begin{cases} d x = \frac{d t}{1 + t^{2}} \\ \sin 2 x = \frac{2 \tan x}{1 + \tan^{2} x} = \frac{2 t}{1 + t^{2}} \end{cases}]$ $T = \int \frac{1 + t^{2} - 2 t}{{(1 + t^{2} + 2 t)}^{2}} d t$ $= \int \frac{{(t - 1)}^{2}}{{(t + 1)}^{4}} d t = \int (\frac{1}{{(t + 1)}^{2}} - \frac{4}{{(t + 1)}^{3}} + \frac{4}{{(t + 1)}^{4}}) d t$ $= - \frac{1}{t + 1} + \frac{2}{{(t + 1)}^{2}} - \frac{4}{3 {(t + 1)}^{3}} + c$ $= - \frac{1}{1 + \tan x} + \frac{2}{{(1 + \tan x)}^{2}} - \frac{4}{{(1 + \tan x)}^{3}} + c$
	Commented by peter frank last updated on 07/Apr/22

	$thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum