∫_( 0) ^(π/4) ((sin x+cos x)/(3+sin 2x)) dx =


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Question Number 69735 by mhmd last updated on 27/Sep/19

$\underset{0}{\int^{π / 4}} \frac{\sin x + \cos x}{3 + \sin 2 x} d x =$
Commented by mathmax by abdo last updated on 27/Sep/19
$I=∫_0 ^(π/4) ((sinx+cosx)/(3+sin(2x)))dx ⇒ I =∫_0 ^(π/4) ((sinx+cosx)/(3+2sinx cosx))dx =_(tan((x/2))=t) ∫_0 ^((√2)−1) ((((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 )))/(3+2((2t)/(1+t^2 ))×((1−t^2 )/(1+t^2 ))))((2dt)/(1+t^2 )) =∫_0 ^((√2)−1) ((2t+1−t^2 )/((1+t^2 )^2 {3+((4t(1−t^2 ))/((1+t^2 )^2 ))}))(2dt =∫_0 ^((√2)−1) ((−2t^2 +4t+2)/(3(t^2 +1)^2 +4t−4t^3 ))dt =∫_0 ^((√2)−1) ((−2t^2 +4t+2)/(3(t^4 +2t^2 +1)+4t−4t^3 ))dt = ∫_0 ^((√2)−1) ((−2t^2 +4t+2)/(3t^4 +6t^2 +3+4t−4t^3 ))dt =∫_0 ^((√2)−1) ((−2t^2 +4t+2)/(3t^4 −4t^3 +6t^2 +4t+3))dt the roots of 3t^4 −4t^3 +6t^2 +4t +3 are t_1 =1+1,4142i (complex) t_2 =1−1,4142i(complex) t_3 =−0,3333+0,4714i (complex) t_4 =−0,3333−0,4714i(complex) ⇒ F(t)=((−2t^2 +4t+2)/(3t^4 −4t^3 +6t^2 +4t+3)) =((−2t^2 +4t +2)/((t−t_1 )(t−t_1 ^− )(t−t_2 )(t−t_2 ^− ))) =((−2t^2 +4t+2)/((t^2 −2Re(t_1 )t +∣t_1 ∣^2 )(t^2 −2Re(t_2 )t +∣t_2 ∣))) =((at +b)/(t^2 −2Re(t_1 )t +∣t_1 ∣^2 )) +((bt +d)/(t^2 −2Re(t_2 )t +∣t_2 ∣^2 )) ⇒ ∫ F(t)dt =∫ ((at+b)/(t^2 −2Re(t_1 )t+∣t_1 ∣^2 ))dt +∫ ((bt+d)/(t^2 −2Re(t_2 )t +∣t_2 ∣^2 ))dt ....be continued...$
$I = \int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{3 + s i n (2 x)} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{3 + 2 s i n x c o s x} d x$ $=_{t a n (\frac{x}{2}) = t} \int_{0}^{\sqrt{2} - 1} \frac{\frac{2 t}{1 + t^{2}} + \frac{1 - t^{2}}{1 + t^{2}}}{3 + 2 \frac{2 t}{1 + t^{2}} \times \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int_{0}^{\sqrt{2} - 1} \frac{2 t + 1 - t^{2}}{{(1 + t^{2})}^{2} {3 + \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}} (2 d t$ $= \int_{0}^{\sqrt{2} - 1} \frac{- 2 t^{2} + 4 t + 2}{3 {(t^{2} + 1)}^{2} + 4 t - 4 t^{3}} d t = \int_{0}^{\sqrt{2} - 1} \frac{- 2 t^{2} + 4 t + 2}{3 (t^{4} + 2 t^{2} + 1) + 4 t - 4 t^{3}} d t$ $= \int_{0}^{\sqrt{2} - 1} \frac{- 2 t^{2} + 4 t + 2}{3 t^{4} + 6 t^{2} + 3 + 4 t - 4 t^{3}} d t = \int_{0}^{\sqrt{2} - 1} \frac{- 2 t^{2} + 4 t + 2}{3 t^{4} - 4 t^{3} + 6 t^{2} + 4 t + 3} d t$ $t h e r o o t s o f 3 t^{4} - 4 t^{3} + 6 t^{2} + 4 t + 3 a r e$ $t_{1} = 1 + 1, 4142 i (c o m p l e x)$ $t_{2} = 1 - 1, 4142 i (c o m p l e x)$ $t_{3} = - 0, 3333 + 0, 4714 i (c o m p l e x)$ $t_{4} = - 0, 3333 - 0, 4714 i (c o m p l e x) \Rightarrow$ $F (t) = \frac{- 2 t^{2} + 4 t + 2}{3 t^{4} - 4 t^{3} + 6 t^{2} + 4 t + 3} = \frac{- 2 t^{2} + 4 t + 2}{(t - t_{1}) (t - {\bar{t}}_{1}) (t - t_{2}) (t - {\bar{t}}_{2})}$ $= \frac{- 2 t^{2} + 4 t + 2}{(t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}) (t^{2} - 2 R e (t_{2}) t + ∣ t_{2} ∣)}$ $= \frac{a t + b}{t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}} + \frac{b t + d}{t^{2} - 2 R e (t_{2}) t + ∣ t_{2} ∣^{2}} \Rightarrow$ $\int F (t) d t = \int \frac{a t + b}{t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}} d t + \int \frac{b t + d}{t^{2} - 2 R e (t_{2}) t + ∣ t_{2} ∣^{2}} d t$ $. . . . b e c o n t i n u e d . . .$
	Answered by Kunal12588 last updated on 27/Sep/19
	$I=∫((sin x + cos x)/(3 + sin 2x))dx let sin x − cos x = t ⇒(sin x + cos x) dx = dt t^2 =1−2sin x cos x =1−sin 2x ⇒t^2 −4=−3−sin 2x x→0⇒t⇒−1 x→(π/4)⇒t⇒0 I=∫(dt/(2^2 −t^2 ))=(1/4)log∣((2+t)/(2−t))∣+c ∫_( 0) ^(π/4) ((sin x+cos x)/(3+sin 2x)) dx =(1/4)[log∣((2+t)/(2−t))∣]_(−1) ^0 =(1/4){0−log1+log3}=(1/4)log3$
	$I = \int \frac{s i n x + c o s x}{3 + s i n 2 x} d x$ $l e t s i n x - c o s x = t$ $\Rightarrow (s i n x + c o s x) d x = d t$ $t^{2} = 1 - 2 s i n x c o s x = 1 - s i n 2 x$ $\Rightarrow t^{2} - 4 = - 3 - s i n 2 x$ $x \to 0 \Rightarrow t \Rightarrow - 1$ $x \to \frac{π}{4} \Rightarrow t \Rightarrow 0$ $I = \int \frac{d t}{2^{2} - t^{2}} = \frac{1}{4} l o g ∣ \frac{2 + t}{2 - t} ∣ + c$ $\underset{0}{\int^{π / 4}} \frac{\sin x + \cos x}{3 + \sin 2 x} d x = \frac{1}{4} {[l o g ∣ \frac{2 + t}{2 - t} ∣]}_{- 1}^{0}$ $= \frac{1}{4} {0 - l o g 1 + l o g 3} = \frac{1}{4} l o g 3$
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