a .∫ (dx/(2sin^2 x+3tg^2 x))=? b .∫(( 1+(x)^(1/3) )/(1+(√x)+(x)^(1/3) +(x)^(1/6) ))dx=? c .∫ ((cosx)/(1+cos2x))dx=? d .∫ ((sin^2 x)/((√2)+(√3).cos^2 x))dx=?


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Question Number 58259 by behi83417@gmail.com last updated on 20/Apr/19

$a . \int \frac{dx}{2 \sin^{2} x + 3 {tg}^{2} x} = ?$ $b . \int \frac{1 + \sqrt[3]{x}}{1 + \sqrt{x} + \sqrt[3]{x} + \sqrt[6]{x}} dx = ?$ $c . \int \frac{cosx}{1 + \cos 2 x} dx = ?$ $d . \int \frac{\sin^{2} x}{\sqrt{2} + \sqrt{3} . \cos^{2} x} dx = ?$
Commented by maxmathsup by imad last updated on 21/Apr/19

$y e s s i r i h a v e f o r g o t t e n \sqrt{3} b u t g i v e o p p o r t o n i t y t o d o m e t h o d s o m e t i m e s$ $i f e e l t i r e d b e c a u s e o f t h e w o r k t h a n k s . .$
Commented by behi83417@gmail.com last updated on 21/Apr/19

$d e a r a b d o! t h a n k y o u f o r s o h a r d a n d n i c e w o r k .$ $e x c u s e m e s i r . s o m e t h i n g w e n t w r o n g i n$ $You can't use 'macro parameter character #' in math mode$ $\Rightarrow 2 \sqrt{2} + \sqrt{3} + \sqrt{3} . c o s t, i s r i g h t .$
Commented by maxmathsup by imad last updated on 20/Apr/19

$c) l e t I = \int \frac{c o s x}{1 + c o s (2 x)} d x \Rightarrow I = \int \frac{c o s x}{1 + 2 {c o s}^{2} x - 1} d x$ $= \int \frac{c o s x}{2 {c o s}^{2} x} d x = \frac{1}{2} \int \frac{d x}{c o s x} =_{t a n (\frac{x}{2}) = t} \frac{1}{2} \int \frac{2 d t}{(1 + t^{2}) \frac{1 - t^{2}}{1 + t^{2}}}$ $= \int \frac{d t}{1 - t^{2}} = \int (\frac{1}{1 - t} + \frac{1}{1 + t}) d t = l n ∣ \frac{1 + t}{1 - t} ∣ + c = l n ∣ \frac{1 + t a n (\frac{x}{2})}{1 - t a n (\frac{x}{2})} ∣ + c$
Commented by maxmathsup by imad last updated on 21/Apr/19
$let A =∫ ((sin^2 x)/((√2) +(√3)cos^2 x))dx =∫ (((1−cos(2x))/2)/((√2) +(√3)((1+cos(2x))/2))) dx =∫ ((1−cos(2x))/(2(√2) +(√3) +(√3)cos(2x)))dx =_(2x =t) ∫ ((1−cos(t))/(2(√2) +(√3)+(√3)cos(t))) (dt/2) =(1/2) ∫ ((1−cos(t))/(2(√2) +(√3)cost)) dt =_(tan((t/2))=u) (1/2) ∫ ((1−((1−u^2 )/(1+u^2 )))/(2(√2) +(√3)((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫ ((2u^2 )/((1+u^2 )(2(√2)(1+u^2 )+(√3)−(√3)u^2 ))) du =∫ ((2u^2 )/((1+u^2 )(2(√2) +(√3) +(2(√2)−(√3))u^2 ))) du let decompose F(u) =((2u^2 )/((u^2 +1){au^2 +b})) (with a=2(√2)−(√3) and b=2(√2) +(√3)) F(u) = ((αu +β)/(u^2 +1)) +((cu +d)/(au^2 +b)) we have F(−u)=F(u) ⇒ ((−αu +β)/(u^2 +1)) +((−cu +d)/(au^2 +b)) =F(u) ⇒α=c =0 ⇒F(u) =(β/(u^2 +1)) +(d/(au^2 +b)) F(0) =0=β +(d/b) ⇒bβ +d =0 lim_(u→0) u^2 F(u) =(2/a) = β +(d/a) ⇒2 =βa +d ⇒d=2−βa ⇒ bβ +2−βa =0 ⇒(b−a)β =−2 ⇒β =(2/(a−b)) d =−bβ =((−2b)/(a−b)) ⇒F(u) =(2/((a−b)(u^2 +1))) −((2b)/((a−b)(au^2 +b))) ⇒ ∫ F(u)du =(2/(a−b)) ∫ (du/(u^2 +1)) −((2b)/((a−b))) ∫ (du/(au^2 +b)) ∫ (du/(u^2 +1)) =arctan(u) +c_0 ∫ (du/(au^2 +b)) =(1/a) ∫ (du/(u^2 +(b/a))) =_(u=(√(b/a))u) (1/a) ∫ (du/((b/a)(1 +u^2 ))) (√(b/a))du =(1/(√(ab))) arctan(u) +c_1 ⇒ A =(2/(a−b)) arctan(tanx) +(1/(√(ab))) actan(tanx) +c =((2x)/(a−b)) +(x/(√(ab))) =((2x)/(−2(√3))) +(x/(.(√5))) =((1/(√5)) −(1/(√3)))x$
$l e t A = \int \frac{{s i n}^{2} x}{\sqrt{2} + \sqrt{3} {c o s}^{2} x} d x = \int \frac{\frac{1 - c o s (2 x)}{2}}{\sqrt{2} + \sqrt{3} \frac{1 + c o s (2 x)}{2}} d x$ $= \int \frac{1 - c o s (2 x)}{2 \sqrt{2} + \sqrt{3} + \sqrt{3} c o s (2 x)} d x =_{2 x = t} \int \frac{1 - c o s (t)}{2 \sqrt{2} + \sqrt{3} + \sqrt{3} c o s (t)} \frac{d t}{2}$ $= \frac{1}{2} \int \frac{1 - c o s (t)}{2 \sqrt{2} + \sqrt{3} c o s t} d t =_{t a n (\frac{t}{2}) = u} \frac{1}{2} \int \frac{1 - \frac{1 - u^{2}}{1 + u^{2}}}{2 \sqrt{2} + \sqrt{3} \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= \int \frac{2 u^{2}}{(1 + u^{2}) (2 \sqrt{2} (1 + u^{2}) + \sqrt{3} - \sqrt{3} u^{2})} d u$ $= \int \frac{2 u^{2}}{(1 + u^{2}) (2 \sqrt{2} + \sqrt{3} + (2 \sqrt{2} - \sqrt{3}) u^{2})} d u l e t d e c o m p o s e$ $F (u) = \frac{2 u^{2}}{(u^{2} + 1) {{a u}^{2} + b}} (w i t h a = 2 \sqrt{2} - \sqrt{3} a n d b = 2 \sqrt{2} + \sqrt{3})$ $F (u) = \frac{α u + β}{u^{2} + 1} + \frac{c u + d}{{a u}^{2} + b} w e h a v e F (- u) = F (u) \Rightarrow$ $\frac{- α u + β}{u^{2} + 1} + \frac{- c u + d}{{a u}^{2} + b} = F (u) \Rightarrow α = c = 0 \Rightarrow F (u) = \frac{β}{u^{2} + 1} + \frac{d}{{a u}^{2} + b}$ $F (0) = 0 = β + \frac{d}{b} \Rightarrow b β + d = 0$ ${l i m}_{u \to 0} u^{2} F (u) = \frac{2}{a} = β + \frac{d}{a} \Rightarrow 2 = β a + d \Rightarrow d = 2 - β a \Rightarrow$ $b β + 2 - β a = 0 \Rightarrow (b - a) β = - 2 \Rightarrow β = \frac{2}{a - b}$ $d = - b β = \frac{- 2 b}{a - b} \Rightarrow F (u) = \frac{2}{(a - b) (u^{2} + 1)} - \frac{2 b}{(a - b) ({a u}^{2} + b)} \Rightarrow$ $\int F (u) d u = \frac{2}{a - b} \int \frac{d u}{u^{2} + 1} - \frac{2 b}{(a - b)} \int \frac{d u}{{a u}^{2} + b}$ $\int \frac{d u}{u^{2} + 1} = a r c t a n (u) + c_{0}$ $\int \frac{d u}{{a u}^{2} + b} = \frac{1}{a} \int \frac{d u}{u^{2} + \frac{b}{a}} =_{u = \sqrt{\frac{b}{a}} u} \frac{1}{a} \int \frac{d u}{\frac{b}{a} (1 + u^{2})} \sqrt{\frac{b}{a}} d u$ $= \frac{1}{\sqrt{a b}} a r c t a n (u) + c_{1} \Rightarrow$ $A = \frac{2}{a - b} a r c t a n (t a n x) + \frac{1}{\sqrt{a b}} a c t a n (t a n x) + c$ $= \frac{2 x}{a - b} + \frac{x}{\sqrt{a b}} = \frac{2 x}{- 2 \sqrt{3}} + \frac{x}{. \sqrt{5}} = (\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{3}}) x$
	Answered by tanmay last updated on 20/Apr/19

	$c) \int \frac{c o s x}{2 {c o s}^{2} x} d x$ $\frac{1}{2} \int s e c x d x$ $\frac{1}{2} l n (s e c x + t a n x) + c$
	Answered by tanmay last updated on 20/Apr/19

	$b) x = t^{6} \to d x = 6 t^{5} d t$ $\int \frac{1 + t^{2}}{1 + t^{3} + t^{2} + t} \times 6 t^{5} d t$ $6 \int \frac{t^{7} + t^{5}}{t^{3} + t^{2} + t + 1} d t$ $= 6 \int \frac{t^{7} + t^{6} + t^{5} + t^{4} - t^{6} - t^{4}}{t^{3} + t^{2} + t + 1} d t$ $= 6 \int t^{4} - 6 \int \frac{t^{6} + t^{4}}{t^{2} (t + 1) + 1 (t^{2} + 1)} d t$ $6 \int t^{4} - 6 \int \frac{t^{4} (t^{2} + 1)}{(t^{2} + 1) (t + 1)} d t$ $= 6 \int t^{4} - 6 \int \frac{t^{4} - 1 + 1}{t + 1} d t$ $6 \int t^{4} - 6 \int \frac{(t^{2} + 1) (t - 1) (t + 1) + 1}{t + 1} d t$ $6 \int t^{4} - 6 \int (t^{2} + 1) (t - 1) d t - 6 \int \frac{d t}{t + 1}$ $6 \int t^{4} - 6 \int t^{3} - t^{2} + t - 1 d t - 6 \int \frac{d t}{t + 1}$ $= \frac{6 t^{5}}{5} - \frac{6 t^{4}}{4} + \frac{6 t^{3}}{3} - \frac{6 t^{2}}{2} + 6 t - 6 l n (t + 1) + c$ $= \frac{6}{5} {(x)}^{\frac{5}{6}} - \frac{3}{2} {(x)}^{\frac{4}{6}} + 2 {(x)}^{\frac{3}{6}} - 3 {(x)}^{\frac{2}{6}} + 6 {(x)}^{\frac{1}{6}} - 6 l n (x^{\frac{1}{6}} + 1) + c$
	Answered by tanmay last updated on 20/Apr/19

	$a) \int \frac{d x}{1 - c o s 2 x + 3 {t a n}^{2} x}$ $t = t a n x \to d t = {s e c}^{2} x d x$ $d x = \frac{d t}{1 + t^{2}}$ $\int \frac{d t}{(1 + t^{2}) (1 - \frac{1 - t^{2}}{1 + t^{2}} + 3 t^{2})}$ $\int \frac{d t}{1 + t^{2} - 1 + t^{2} + 3 t^{2} + 3 t^{4}}$ $\int \frac{d t}{3 t^{4} + 5 t^{2}}$ $\int \frac{d t}{t^{2} (3 t^{2} + 5)}$ $\frac{1}{5} \int \frac{3 t^{2} + 5 - 3 t^{2}}{t^{2} (3 t^{2} + 5)} d t$ $\frac{1}{5} \int \frac{d t}{t^{2}} - \frac{3}{5} \int \frac{d t}{3 (t^{2} + \frac{5}{3})}$ $\frac{1}{5} \times \frac{t^{- 2 + 1}}{- 2 + 1} - \frac{1}{5} \times \frac{1}{\sqrt{\frac{5}{3}}} \times {t a n}^{- 1} (\frac{t}{\sqrt{\frac{5}{3}}}) + c$ $\frac{- 1}{5 t} - \frac{\sqrt{3}}{5 \sqrt{5}} {t a n}^{- 1} (\frac{t}{\sqrt{\frac{5}{3}}}) + c$ $\frac{- 1}{5 t a n x} - \frac{\sqrt{3}}{5 \sqrt{5}} {t a n}^{- 1} (\frac{t a n x}{\sqrt{\frac{5}{3}}}) + c$
	Answered by tanmay last updated on 20/Apr/19

	$4) \int \frac{{s i n}^{2} x}{a + {b c o s}^{2} x} d x$ $\int \frac{\frac{1 - c o s 2 x}{2}}{a + b (\frac{1 + c o s 2 x}{2})} d x$ $\int \frac{1 - c o s 2 x}{2 a + b + b c o s 2 x} d x$ $\frac{1}{b} \int \frac{1 - c o s 2 x}{(\frac{2 a + b}{b}) + c o s 2 x} d x$ $\frac{1}{b} \int \frac{1 + (\frac{2 a + b}{b}) - (\frac{2 a + b}{b}) - c o s 2 x}{(\frac{2 a + b}{b}) + c o s 2 x} d x$ $l e t k = \frac{2 a + b}{b}$ $\frac{1}{b} \int \frac{1 + k}{k + c o s 2 x} d x - \frac{1}{b} \int d x$ $\frac{1 + k}{b} \int \frac{d x}{k + \frac{1 - {t a n}^{2} x}{1 + {t a n}^{2} x}} - \frac{1}{b} \int d x$ $\frac{1 + k}{b} \int \frac{{s e c}^{2} x d x}{(k + 1) + (k - 1) {t a n}^{2} x} - \frac{1}{b} \int d x$ $\frac{1 + k}{b} \times \frac{1}{k - 1} \int \frac{d (t a n x)}{{(\sqrt{\frac{k + 1}{k - 1}})}^{2} + {t a n}^{2} x} - \frac{1}{b} \int d x$ $\frac{1}{b} \times \frac{k + 1}{k - 1} \times \frac{1}{\sqrt{\frac{k + 1}{k - 1}}} {t a n}^{- 1} (\frac{t a n x}{\sqrt{\frac{k + 1}{k - 1}}}) - \frac{1}{b} x + c$ $n o w a = \sqrt{2} b = \sqrt{3} k = \frac{2 a + b}{b} \to k = \frac{2 \sqrt{2} + \sqrt{3}}{\sqrt{3}}$ $\frac{1}{\sqrt{3}} \times \sqrt{\frac{\frac{2 \sqrt{2} + \sqrt{3}}{\sqrt{3}} + 1}{\frac{2 \sqrt{2} + \sqrt{3}}{\sqrt{3}} - 1}} \times {t a n}^{- 1} (\frac{t a n x}{\sqrt{\frac{\sqrt{3} + \sqrt{2}}{\sqrt{2}}}}) - \frac{1}{\sqrt{3}} x + c$ $\frac{1}{\sqrt{3}} \times \sqrt{\frac{\sqrt{3} + \sqrt{2}}{\sqrt{2}}} \times {t a n}^{- 1} (\frac{t a n x}{\sqrt{\frac{\sqrt{3} + \sqrt{2}}{\sqrt{2}}}}) - \frac{1}{\sqrt{3}} x + c$
	Commented by behi83417@gmail.com last updated on 20/Apr/19

	$s i r t a n m a y! t h a n k y o u v e r y m u c h f o r$ $s o h a r d w o r k . n i c e a n d s m a r t .$
	Commented by peter frank last updated on 20/Apr/19

	$n i c e w o r k . t h a n k s$
	Commented by tanmay last updated on 21/Apr/19

	$t h a n k y o u s i r$
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