Evaluate-1-2-x-4-x-dx-2-x-2-x-4-dx-3-x-2-x-4-dx-4-dx-2sinx-3secx-

Question Number 53119 by rahul 19 last updated on 18/Jan/19

E v a l u a t e :

1) \int \sqrt{\frac{2 - x}{4 + x}} d x

2) \int \sqrt{\frac{x - 2}{x - 4}} d x

3) \int \sqrt{(x - 2) (x - 4)} d x

4) \int \frac{d x}{2 \sin \boldsymbol x + 3 \sec \boldsymbol x} .

Commented by maxmathsup by imad last updated on 18/Jan/19

3) let I =∫(√((x−2)(x−4)))dx changement (√(x−2))=t give x−2=t^2 ⇒x=2+t^2 I =∫ t(√(2+t^2 −4))2tdt =2 ∫ t^2 (√(t^2 −2))dt =_(t=(√2)ch(u)) 2 ∫ 2ch^2 t (√2)sh(u)(√2)sh(u) du =8 ∫ ch^2 u sh^2 u du =8 ∫ (((2shu chu)^2 )/4) du =2 ∫ sh^2 u du =2 ∫ ((ch(2u)−1)/2) du =∫ ch(2u)du −u +c =(1/2)sh(2u) −u +c but u=argch((t/( (√2)))) =ln((t/( (√2))) +(√((t^2 /2)−1))) ⇒ sh(2u) =((e^(2u) −e^(−2u) )/2) =(1/2){((t/( (√2))) +(√((t^2 /2)−1)))^2 −((t/( (√2))) +(√((t^2 /2)−1)))^(−2) } ⇒ I =(1/4){ (((√(x−2))/( (√2))) +(√(((x−2)/2)−1)))^2 −(((√(x−2))/( (√2)))+(√(((x−2)/2)−1)))^(−2) −ln(((√(x−2))/( (√2))) +(√(((x−2)/2)−1))) +c =(1/4){ (1/2)((√(x−2))+(√(x−4)))^2 −(2/(((√(x−2))+(√(x−4)))^2 ))}−ln((((√(x−2))+(√(x−4)))/( (√2)))) +c

3) l e t I = \int \sqrt{(x - 2) (x - 4)} d x c h a n g e m e n t \sqrt{x - 2} = t g i v e x - 2 = t^{2} \Rightarrow x = 2 + t^{2}

I = \int t \sqrt{2 + t^{2} - 4} 2 t d t = 2 \int t^{2} \sqrt{t^{2} - 2} d t =_{t = \sqrt{2} c h (u)} 2 \int 2 {c h}^{2} t \sqrt{2} s h (u) \sqrt{2} s h (u) d u

= 8 \int {c h}^{2} u {s h}^{2} u d u = 8 \int \frac{{(2 s h u c h u)}^{2}}{4} d u = 2 \int {s h}^{2} u d u

= 2 \int \frac{c h (2 u) - 1}{2} d u = \int c h (2 u) d u - u + c

= \frac{1}{2} s h (2 u) - u + c b u t u = a r g c h (\frac{t}{\sqrt{2}}) = l n (\frac{t}{\sqrt{2}} + \sqrt{\frac{t^{2}}{2} - 1}) \Rightarrow

s h (2 u) = \frac{e^{2 u} - e^{- 2 u}}{2} = \frac{1}{2} {{(\frac{t}{\sqrt{2}} + \sqrt{\frac{t^{2}}{2} - 1})}^{2} - {(\frac{t}{\sqrt{2}} + \sqrt{\frac{t^{2}}{2} - 1})}^{- 2}} \Rightarrow

I = \frac{1}{4} {{(\frac{\sqrt{x - 2}}{\sqrt{2}} + \sqrt{\frac{x - 2}{2} - 1})}^{2} - {(\frac{\sqrt{x - 2}}{\sqrt{2}} + \sqrt{\frac{x - 2}{2} - 1})}^{- 2} - l n (\frac{\sqrt{x - 2}}{\sqrt{2}} + \sqrt{\frac{x - 2}{2} - 1}) + c

= \frac{1}{4} {\frac{1}{2} {(\sqrt{x - 2} + \sqrt{x - 4})}^{2} - \frac{2}{{(\sqrt{x - 2} + \sqrt{x - 4})}^{2}}} - l n (\frac{\sqrt{x - 2} + \sqrt{x - 4}}{\sqrt{2}}) + c

Commented by Tawa1 last updated on 18/Jan/19

Sir please check your solution to my question 52841. i asked for some clarification sir. Thanks sir

Sir please check your solution to my question 52841 .

i asked for some clarification sir . Thanks sir

Commented by maxmathsup by imad last updated on 23/Jan/19

2) let I =∫(√((x−2)/(x−4)))dx changement (√((x−2)/(x−4)))=t give ((x−2)/(x−4)) =t^2 ⇒ x−2 =t^2 x−4t^2 ⇒(1−t^2 )x =2−4t^2 ⇒x =((−4t^2 +2)/(1−t^2 )) =((4t^2 −2)/(t^2 −1)) ⇒ (dx/dt)=((8t(t^2 −1)−(4t^2 −2)2t)/((t^2 −1)^2 )) =((8t^3 −8t −8t^3 +4t)/((t^2 −1)^2 )) =((−4t)/((t^2 −1)^2 )) ⇒ I = ∫ t(((−4t)/((t^2 −1)^2 )))dt =−4 ∫ (t^2 /((t^2 −1)^2 ))dt =−4∫((t^2 −1 +1)/((t^2 −1)^2 ))dt =−4 ∫ (dt/(t^2 −1)) −4 ∫ (dt/((t^2 −1)^2 )) but ∫ (dt/(t^2 −1)) =(1/2)∫ ((1/(t−1)) −(1/(t+1)))dt =(1/2)ln∣((t−1)/(t+1))∣ +c_1 let decompose F(t)=(1/((t^2 −1)^2 )) ⇒F(t)=(1/((t−1)^2 (t+1)^2 )) =(a/(t−1)) +(b/((t−1)^2 )) +(c/((t+1))) +(d/((t+1)^2 )) b =lim_(t→1) (t−1)^2 F(t) =(1/4) d =lim_(t→−1) (t+1)^2 F(t) =(1/4) ⇒F(t)=(a/(t−1)) +(1/(4(t−1)^2 )) +(c/(t+1)) +(1/(4(t+1)^2 )) F(−t)=F(t) ⇒((−a)/(t+1)) +(1/(4(t+1)^2 )) + ((−c)/(t−1)) +(1/(4(t−1)^2 )) =F(t) ⇒c=−a ⇒ F(t) =(a/(t−1)) +(1/(4(t−1)^2 )) −(a/(t+1)) +(1/(4(t+1)^2 )) F(0)=1 =−a +(1/4) −a +(1/4) ⇒−2a+(1/2) =1 ⇒−2a =(1/2) ⇒a =−(1/4) ⇒ F(x)=−(1/(4(t−1))) +(1/(4(t+1))) +(1/(4(t−1)^2 )) +(1/(4(t+1)^2 )) ⇒ ∫ (dt/((t^2 −1)^2 )) =(1/4)ln∣((t+1)/(t−1))∣−(1/(4(t−1))) −(1/(4(t+1))) +c_2 ⇒ I =−2ln∣((t−1)/(t+1))∣ −ln∣((t+1)/(t−1))∣ +(1/(t−1)) +(1/(t+1)) +C ⇒ I =ln∣((t+1)/(t−1))∣ +(1/(t−1)) +(1/(t+1)) +C I =ln∣(((√((x−2)/(x−4)))+1)/( (√((x−2)/(x−4)))−1))∣ +(1/( (√((x−2)/(x−4)))−1)) +(1/( (√((x−2)/(x−4)))+1)) +C .

2) l e t I = \int \sqrt{\frac{x - 2}{x - 4}} d x c h a n g e m e n t \sqrt{\frac{x - 2}{x - 4}} = t g i v e \frac{x - 2}{x - 4} = t^{2} \Rightarrow

x - 2 = t^{2} x - 4 t^{2} \Rightarrow (1 - t^{2}) x = 2 - 4 t^{2} \Rightarrow x = \frac{- 4 t^{2} + 2}{1 - t^{2}} = \frac{4 t^{2} - 2}{t^{2} - 1} \Rightarrow

\frac{d x}{d t} = \frac{8 t (t^{2} - 1) - (4 t^{2} - 2) 2 t}{{(t^{2} - 1)}^{2}} = \frac{8 t^{3} - 8 t - 8 t^{3} + 4 t}{{(t^{2} - 1)}^{2}} = \frac{- 4 t}{{(t^{2} - 1)}^{2}} \Rightarrow

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

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

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

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

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

b = {l i m}_{t \to 1} {(t - 1)}^{2} F (t) = \frac{1}{4}

d = {l i m}_{t \to - 1} {(t + 1)}^{2} F (t) = \frac{1}{4} \Rightarrow F (t) = \frac{a}{t - 1} + \frac{1}{4 {(t - 1)}^{2}} + \frac{c}{t + 1} + \frac{1}{4 {(t + 1)}^{2}}

F (- t) = F (t) \Rightarrow \frac{- a}{t + 1} + \frac{1}{4 {(t + 1)}^{2}} + \frac{- c}{t - 1} + \frac{1}{4 {(t - 1)}^{2}} = F (t) \Rightarrow c = - a \Rightarrow

F (t) = \frac{a}{t - 1} + \frac{1}{4 {(t - 1)}^{2}} - \frac{a}{t + 1} + \frac{1}{4 {(t + 1)}^{2}}

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

F (x) = - \frac{1}{4 (t - 1)} + \frac{1}{4 (t + 1)} + \frac{1}{4 {(t - 1)}^{2}} + \frac{1}{4 {(t + 1)}^{2}} \Rightarrow

\int \frac{d t}{{(t^{2} - 1)}^{2}} = \frac{1}{4} l n ∣ \frac{t + 1}{t - 1} ∣ - \frac{1}{4 (t - 1)} - \frac{1}{4 (t + 1)} + c_{2} \Rightarrow

I = - 2 l n ∣ \frac{t - 1}{t + 1} ∣ - l n ∣ \frac{t + 1}{t - 1} ∣ + \frac{1}{t - 1} + \frac{1}{t + 1} + C \Rightarrow

I = l n ∣ \frac{t + 1}{t - 1} ∣ + \frac{1}{t - 1} + \frac{1}{t + 1} + C

I = l n ∣ \frac{\sqrt{\frac{x - 2}{x - 4}} + 1}{\sqrt{\frac{x - 2}{x - 4}} - 1} ∣ + \frac{1}{\sqrt{\frac{x - 2}{x - 4}} - 1} + \frac{1}{\sqrt{\frac{x - 2}{x - 4}} + 1} + C .

Answered by MJS last updated on 18/Jan/19

(1) t=(√((4+x)/(2−x))) → x=((2(t^2 −2))/(t^2 +1)); dx=((√((2−x)^3 (4+x)))/3)dt 12∫(dt/((t^2 +1)^2 )) u=arctan t → t=tan u; dt=(t^2 +1)du 12∫cos^2 u du

(1)

t = \sqrt{\frac{4 + x}{2 - x}} \to x = \frac{2 (t^{2} - 2)}{t^{2} + 1}; d x = \frac{\sqrt{{(2 - x)}^{3} (4 + x)}}{3} d t

12 \int \frac{d t}{{(t^{2} + 1)}^{2}}

u = \arctan t \to t = \tan u; d t = (t^{2} + 1) d u

12 \int \cos^{2} u d u

Answered by MJS last updated on 18/Jan/19

(2) t=(√((x−4)/(x−2))) → x=((2(t^2 −2))/((t^2 −1))); dx=(√((x−4)(x−2)^3 ))dt 4∫(dt/((t^2 −1)^2 )) u=arcsin (1/t) → t=(1/(sin u)); dt=−t(√(t^2 −1))du −4∫((sin^2 u)/(cos^3 u))du=−4∫sec^3 u −sec u du

(2)

t = \sqrt{\frac{x - 4}{x - 2}} \to x = \frac{2 (t^{2} - 2)}{(t^{2} - 1)}; d x = \sqrt{(x - 4) {(x - 2)}^{3}} d t

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

u = \arcsin \frac{1}{t} \to t = \frac{1}{\sin u}; d t = - t \sqrt{t^{2} - 1} d u

- 4 \int \frac{\sin^{2} u}{\cos^{3} u} d u = - 4 \int \sec^{3} u - \sec u d u

Answered by MJS last updated on 18/Jan/19

(3) t=x−3 → dt=dx ∫(√(t^2 −1))dt u=arccos (1/t) → t=(1/(cos u)); dt=t(√(t^2 −1))du ∫((sin^2 u)/(cos^3 u))du ...

(3)

t = x - 3 \to d t = d x

\int \sqrt{t^{2} - 1} d t

u = \arccos \frac{1}{t} \to t = \frac{1}{\cos u}; d t = t \sqrt{t^{2} - 1} d u

\int \frac{\sin^{2} u}{\cos^{3} u} d u \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

1)∫((2−x)/( (√((4+x)(2−x)))))dx ∫((2−x)/( (√(8−4x+2x−x^2 ))))dx ∫((2−x)/( (√(8−2x−x^2 )))) (1/2)∫((−2−2x+6)/( (√(8−2x−x^2 ))))dx =(1/2)∫((d(8−2x−x^2 ))/( (√(8−2x−x^2 ))))+3∫(dx/( (√((3)^2 −(x+1)^2 )))) =(1/2)×(((8−2x−x^2 )^(((−1)/2)+1) )/(((−1)/2)+1))+3sin^(−1) (((x+1)/3))+c =(8−2x−x^2 )^(1/2) +3sin^(−1) (((x+1)/3))+c 2)∫((x−2)/( (√((x−2)(x−4)))))dx ∫((x−2)/( (√(x^2 −6x+8))))dx (1/2)∫((2x−6+2)/( (√(x^2 −6x+8))))dx (1/2)∫((d(x^2 −6x+8))/( (√(x^2 −6x+8))))+∫(dx/( (√((x−3)^2 −1)))) (1/2)×(((x^2 −6x+8)^(((−1)/2)+1) )/(((−1)/2)+1))+ln{(x−3)+(√((x−3)^2 −1)) }+c =(x^2 −6x+8)^(1/2) +ln{(x−3)+(√(x^2 −6x+8)) }+c 3∫(√((x−2)(x−4))) dx ∫(√((x−3)^2 −1)) dx =(((x−3))/2)(√((x−3)^2 −1)) −(1/2)ln{(x−3)+(√((x−3)^2 −1)) }+c =(((x−3))/2)(√(x^2 −6x+8)) −(1/2)ln{(x−3)+(√((x^2 −6x+8))

1) \int \frac{2 - x}{\sqrt{(4 + x) (2 - x)}} d x

\int \frac{2 - x}{\sqrt{8 - 4 x + 2 x - x^{2}}} d x

\int \frac{2 - x}{\sqrt{8 - 2 x - x^{2}}}

\frac{1}{2} \int \frac{- 2 - 2 x + 6}{\sqrt{8 - 2 x - x^{2}}} d x

= \frac{1}{2} \int \frac{d (8 - 2 x - x^{2})}{\sqrt{8 - 2 x - x^{2}}} + 3 \int \frac{d x}{\sqrt{{(3)}^{2} - {(x + 1)}^{2}}}

= \frac{1}{2} \times \frac{{(8 - 2 x - x^{2})}^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + 3 {s i n}^{- 1} (\frac{x + 1}{3}) + c

= {(8 - 2 x - x^{2})}^{\frac{1}{2}} + 3 {s i n}^{- 1} (\frac{x + 1}{3}) + c

2) \int \frac{x - 2}{\sqrt{(x - 2) (x - 4)}} d x

\int \frac{x - 2}{\sqrt{x^{2} - 6 x + 8}} d x

\frac{1}{2} \int \frac{2 x - 6 + 2}{\sqrt{x^{2} - 6 x + 8}} d x

\frac{1}{2} \int \frac{d (x^{2} - 6 x + 8)}{\sqrt{x^{2} - 6 x + 8}} + \int \frac{d x}{\sqrt{{(x - 3)}^{2} - 1}}

\frac{1}{2} \times \frac{{(x^{2} - 6 x + 8)}^{\frac{- 1}{2} + 1}}{\frac{- 1}{2} + 1} + l n {(x - 3) + \sqrt{{(x - 3)}^{2} - 1}} + c

= {(x^{2} - 6 x + 8)}^{\frac{1}{2}} + l n {(x - 3) + \sqrt{x^{2} - 6 x + 8}} + c

3 \int \sqrt{(x - 2) (x - 4)} d x

\int \sqrt{{(x - 3)}^{2} - 1} d x

= \frac{(x - 3)}{2} \sqrt{{(x - 3)}^{2} - 1} - \frac{1}{2} l n {(x - 3) + \sqrt{{(x - 3)}^{2} - 1}} + c

= \frac{(x - 3)}{2} \sqrt{x^{2} - 6 x + 8} - \frac{1}{2} l n {(x - 3) + \sqrt{(x^{2} - 6 x + 8}

Commented by rahul 19 last updated on 18/Jan/19

T h a n k y o u S i r!

Commented by Otchere Abdullai last updated on 18/Jan/19

w o w! w e l l d o n e s i r!

Answered by MJS last updated on 18/Jan/19

(4) =∫((cos x)/(3+cos x sin x))dx Weyerstrass t=tan (x/2) → x=2arctan t; dx=2cos^2 (x/2) dt=((2dt)/(t^2 +1)) cos x =−((t^2 −1)/(t^2 +1)); sin x =((2t)/(t^2 +1)) −2∫((t^2 −1)/((t^2 −2t+3)(3t^2 +2t+1)))dt now decompose... ((t^2 −1)/((t^2 −2t+3)(3t^2 +2t+1)))=(1/8)×((t+1)/(t^2 −2t+3))−(3/8)×((t+1)/(3t^2 +2t+1)) ((t+1)/(t^2 −2t+3))=((t−1)/(t^2 −2t+3))+(2/(t^2 −2t+3)) ((t+1)/(3t^2 +2t+1))=((6t+2)/(6(3t^2 +2t+1)))+(4/(6(3t^2 +2t+1)))= =(1/3)×((3t+1)/(3t^2 +2t+1))+(2/3)×(1/(3t^2 +2t+1)) now use formulas

(4)

= \int \frac{\cos x}{3 + \cos x \sin x} d x

Weyerstrass

t = \tan \frac{x}{2} \to x = 2 \arctan t; d x = {2 \cos}^{2} \frac{x}{2} d t = \frac{2 d t}{t^{2} + 1}

\cos x = - \frac{t^{2} - 1}{t^{2} + 1}; \sin x = \frac{2 t}{t^{2} + 1}

- 2 \int \frac{t^{2} - 1}{(t^{2} - 2 t + 3) (3 t^{2} + 2 t + 1)} d t

now decompose \dots

\frac{t^{2} - 1}{(t^{2} - 2 t + 3) (3 t^{2} + 2 t + 1)} = \frac{1}{8} \times \frac{t + 1}{t^{2} - 2 t + 3} - \frac{3}{8} \times \frac{t + 1}{3 t^{2} + 2 t + 1}

\frac{t + 1}{t^{2} - 2 t + 3} = \frac{t - 1}{t^{2} - 2 t + 3} + \frac{2}{t^{2} - 2 t + 3}

\frac{t + 1}{3 t^{2} + 2 t + 1} = \frac{6 t + 2}{6 (3 t^{2} + 2 t + 1)} + \frac{4}{6 (3 t^{2} + 2 t + 1)} =

= \frac{1}{3} \times \frac{3 t + 1}{3 t^{2} + 2 t + 1} + \frac{2}{3} \times \frac{1}{3 t^{2} + 2 t + 1}

now use formulas

Commented by rahul 19 last updated on 18/Jan/19

T h a n k Y o u S i r!

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jan/19

4)∫(dx/(2sinx+3secx)) ∫((cosxdx)/(3+2sinxcosx)) (1/2)∫((cosx−sinx+cosx+sinx)/(3+2sinxcosx))dx (1/2)∫((cosx−sinx)/(2+(sinx+cosx)^2 ))dx+(1/2)∫((d(sinx−cosx))/(4−(1−2sinxcosx))) (1/2)∫((d(sinx+cosx))/(((√2) )^2 +(sinx+cosx)^2 ))+(1/2)∫((d(sinx−cosx))/(4−(sinx−cosx)^2 ))←formula∫(dx/(a^2 −x^2 )) (1/2)×(1/( (√2)))tan^(−1) (((sinx+cosx)/( (√2))))+(1/2)×(1/(2×2))ln(((2+sinx−cosx)/(2−sinx+cosx)))+c

4) \int \frac{d x}{2 s i n x + 3 s e c x}

\int \frac{c o s x d x}{3 + 2 s i n x c o s x}

\frac{1}{2} \int \frac{c o s x - s i n x + c o s x + s i n x}{3 + 2 s i n x c o s x} d x

\frac{1}{2} \int \frac{c o s x - s i n x}{2 + {(s i n x + c o s x)}^{2}} d x + \frac{1}{2} \int \frac{d (s i n x - c o s x)}{4 - (1 - 2 s i n x c o s x)}

\frac{1}{2} \int \frac{d (s i n x + c o s x)}{{(\sqrt{2})}^{2} + {(s i n x + c o s x)}^{2}} + \frac{1}{2} \int \frac{d (s i n x - c o s x)}{4 - {(s i n x - c o s x)}^{2}} \leftarrow f o r m u l a \int \frac{d x}{a^{2} - x^{2}}

\frac{1}{2} \times \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{s i n x + c o s x}{\sqrt{2}}) + \frac{1}{2} \times \frac{1}{2 \times 2} l n (\frac{2 + s i n x - c o s x}{2 - s i n x + c o s x}) + c

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Evaluate-1-2-x-4-x-dx-2-x-2-x-4-dx-3-x-2-x-4-dx-4-dx-2sinx-3secx-

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