How-can-solve-tan-x-dx-

Question Number 55520 by Rdk96 last updated on 26/Feb/19

H o w c a n s o l v e \int \sqrt{} \tan (x) d x ?

Commented by maxmathsup by imad last updated on 26/Feb/19

let A =∫ (√(tanx))dx changement (√(tanx))=t give tanx =t^2 ⇒x =arctan(t^2 ) ⇒ A =∫ t ((2t)/(1+t^4 )) dt =2 ∫ (t^2 /(t^4 +1))dt let decompose F(t)=(t^2 /(t^4 +1)) ⇒ F(t)=(t^2 /((t^2 −(√2)t +1)(t^2 +(√2)t +1))) =((at +b)/(t^2 −(√2)t +1)) +((ct +d)/(t^2 +(√2)t +1)) F(−t)=F(t) ⇒((−at +b)/(t^2 +(√2)t +1))+((−ct +d)/(t^2 −(√2)t +1)) =F(t) ⇒c=−a and d=b ⇒ F(t)=((at +b)/(t^2 −(√2)t+1)) +((−at +b)/(t^2 +(√2)t +1)) F(0) =0 =2b ⇒b=0 ⇒F(t)=((at)/(t^2 −(√2)t +1)) −((at)/(t^2 +(√2)t +1)) F(1) =(1/((2−(√2))(2+(√2)))) = (a/(2−(√2))) −(a/(2+(√2))) ⇒(1/2) =(((2+(√2))a−(2−(√2))a)/2) ⇒ 2(√2)a =1 ⇒a =(1/(2(√2))) ⇒F(t)=(1/(2(√2))){ (t/(t^2 −(√2)t +1)) −(t/(t^2 +(√2)t +1))} ⇒ A =(1/( (√2))){ ∫ ((tdt)/(t^2 −(√2)t +1)) −∫ ((tdt)/(t^2 +(√2)t +1))} +c but ∫ ((tdt)/(t^2 −(√2)t +1)) =(1/2) ∫ ((2t−(√2)+(√2))/(t^2 −(√2)t +1)) dt =(1/2)ln(t^2 −(√2)t +1)+(1/( (√2)))∫ (dt/(t^2 −(√2)t +1)) ∫ (dt/(t^2 −(√2)t +1)) =∫ (dt/((t−((√2)/2))^2 +(1/2))) =_(t−(1/( (√2)))=(u/( (√2)))) ∫ (du/( (√2)(1/2)(1+u^2 ))) =(√2)arctan(t(√2)−1) ⇒∫ ((tdt)/(t^2 −(√2)t +1)) =(1/2)ln(t^2 −(√2)t +1) +arctan(t(√2)−1) also we get ∫ ((tdt)/(t^2 +(√2)t +1)) =(1/2)ln(t^2 +(√2)t +1) +arctan(t(√2)+1) ⇒ (√2)A =(1/2)ln(t^2 −(√2)t +1)+arctan(t(√2)−1)−(1/2)ln(t^2 +(√2)t +1)−arctan(t(√2)+1) +c =ln(√((t^2 −(√2)t+1)/(t^2 +(√2)t +1))) +arctan(t(√2)−1)−arctan(t(√2)+1) +c ⇒ A =(1/( (√2))){ ln((√((tanx−(√(2tanx))+1)/(tanx+(√(2tanx))+1)))) +arctan((√(2tanx))−1)−arctan((√(2tanx)) +1)} +c

l e t A = \int \sqrt{t a n x} d x c h a n g e m e n t \sqrt{t a n x} = t g i v e t a n x = t^{2} \Rightarrow x = a r c t a n (t^{2}) \Rightarrow

A = \int t \frac{2 t}{1 + t^{4}} d t = 2 \int \frac{t^{2}}{t^{4} + 1} d t l e t d e c o m p o s e F (t) = \frac{t^{2}}{t^{4} + 1} \Rightarrow

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

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

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

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

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

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

A = \frac{1}{\sqrt{2}} {\int \frac{t d t}{t^{2} - \sqrt{2} t + 1} - \int \frac{t d t}{t^{2} + \sqrt{2} t + 1}} + c b u t

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

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

= \sqrt{2} a r c t a n (t \sqrt{2} - 1) \Rightarrow \int \frac{t d t}{t^{2} - \sqrt{2} t + 1} = \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + a r c t a n (t \sqrt{2} - 1)

a l s o w e g e t \int \frac{t d t}{t^{2} + \sqrt{2} t + 1} = \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) + a r c t a n (t \sqrt{2} + 1) \Rightarrow

\sqrt{2} A = \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + a r c t a n (t \sqrt{2} - 1) - \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) - a r c t a n (t \sqrt{2} + 1) + c

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

A = \frac{1}{\sqrt{2}} {l n (\sqrt{\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}}) + a r c t a n (\sqrt{2 t a n x} - 1) - a r c t a n (\sqrt{2 t a n x} + 1)} + c

Commented by maxmathsup by imad last updated on 26/Feb/19

∫ ((tdt)/(t^2 +(√2)t +1)) =_(t=−u) ∫ (((−u)(−du))/(u^2 −(√2)u +1)) =∫ ((udu)/(u^2 −(√2)u +1)) =(1/2)ln(u^2 −(√2)u +1) +arctan(u(√2)−1) =(1/2)ln(t^2 +(√2)t +1)−arctan(t(√2)+1) ⇒ (√2)A =(1/2)ln(t^2 −(√2)t +1)+arctan(t(√2)−1)−(1/2)ln(t^2 +(√2)t +1) +arctan(t(√2)+1) A =(1/( (√2))){ ln((√((tanx−2(√(tanx))+1)/(tanx +(√(2tanx))+1)))) +arctan(t(√2)−1) +arctan(t(√2)+1)} +c .

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

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

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

\sqrt{2} A = \frac{1}{2} l n (t^{2} - \sqrt{2} t + 1) + a r c t a n (t \sqrt{2} - 1) - \frac{1}{2} l n (t^{2} + \sqrt{2} t + 1) + a r c t a n (t \sqrt{2} + 1)

A = \frac{1}{\sqrt{2}} {l n (\sqrt{\frac{t a n x - 2 \sqrt{t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}}) + a r c t a n (t \sqrt{2} - 1) + a r c t a n (t \sqrt{2} + 1)} + c .

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Feb/19

t^2 =tanx 2tdt=sec^2 xdx ((2tdt)/(1+t^4 ))=dx ∫t×((2tdt)/(1+t^4 )) ∫((2t^2 )/(1+t^4 ))dt ∫((2dt)/(t^2 +(1/t^2 ))) ∫((1−(1/t^2 )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt←main step ∫((1−(1/t^2 ))/((t+(1/t))^2 −2))dt+∫((1+(1/t^2 ))/((t−(1/t))^2 +2))dt ∫((d(t+(1/t)))/((t+(1/t))^2 −2))+∫((d(t−(1/t)))/((t−(1/t))^2 +2)) now use formula ∫(dx/(x^2 −a^2 ))=(1/(2a))ln(((x−a)/(x+a)))+c and∫(dx/(x^2 +a^2 ))=(1/a)tan^(−1) ((x/a))+c_1 (1/(2(√2)))ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))+(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2))))+c (1/(2(√2)))ln((((√(tanx)) +(1/( (√(tanx)) ))−(√2))/( (√(tanx)) +(1/( (√(tanx)) ))+(√2))))+(1/( (√2)))tan^(−1) ((((√(tanx)) −(1/( (√(tanx)) )))/( (√2))))+c pls check...

t^{2} = t a n x 2 t d t = {s e c}^{2} x d x

\frac{2 t d t}{1 + t^{4}} = d x

\int t \times \frac{2 t d t}{1 + t^{4}}

\int \frac{2 t^{2}}{1 + t^{4}} d t

\int \frac{2 d t}{t^{2} + \frac{1}{t^{2}}}

\int \frac{1 - \frac{1}{t^{2}} + 1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t \leftarrow m a i n s t e p

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

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

n o w u s e f o r m u l a \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} l n (\frac{x - a}{x + a}) + c

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

\frac{1}{2 \sqrt{2}} l n (\frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}}) + \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}}) + c

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

p l s c h e c k \dots

Commented by Rdk96 last updated on 26/Feb/19

it is possible to multiply the differential with the equation??

i t i s p o s s i b l e t o m u l t i p l y t h e d i f f e r e n t i a l

w i t h t h e e q u a t i o n ? ?

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Feb/19

yes... ∫(x/(1+x^2 ))dx→(1/2)∫((d(1+x^2 ))/(1+x^2 ))→(1/2)ln(1+x^2 )+c

y e s \dots

\int \frac{x}{1 + x^{2}} d x \to \frac{1}{2} \int \frac{d (1 + x^{2})}{1 + x^{2}} \to \frac{1}{2} l n (1 + x^{2}) + c

Commented by Rdk96 last updated on 26/Feb/19

c o r r e c t .

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Feb/19

or method... p=∫(√(tanx)) +(√(cotx)) dx q=(√(tanx)) −(√(cotx)) dx ∫(√(tanx)) dx=((p+q)/2) p=∫(√((sinx)/(cosx))) +(√((cosx)/(sinx))) dx p=∫((sinx+cosx)/( (√(sinxcosx))))dx p=(√2)∫((d(sinx−cosx))/( (√(1−1+2sinxcosx))))dx =(√2) ∫((d(sinx−cosx))/( (√(1−(sinx−cosx)^2 )) )) =(√2) sin^(−1) (sinx−cosx) +c_1 q=∫((√((sinx)/(cosx))) −(√((cosx)/(sinx))) ) dx q=∫((sinx−cosx)/( (√(sinxcosx))))dx→−∫((cosx−sinx)/( (√(sinxcosx))))dx q=(−(√2) )∫((d(sinx+cosx))/( (√(1+2sinxcosx−1)))) =(−(√2) )∫((d(sinx+cosx))/( (√((sinx+cosx)^2 −1)))) =(−(√2) )ln{(sinx+cosx)+(√((sinx+cosx)^2 −1)) } so ∫(√(tanx)) dx ((√2)/2)[sin^(−1) (sinx−cosx)−ln{(sinx+cosx)+(√((sinx+cosx)^2 −1)) }]+c pls check...

o r m e t h o d \dots

p = \int \sqrt{t a n x} + \sqrt{c o t x} d x

q = \sqrt{t a n x} - \sqrt{c o t x} d x

\int \sqrt{t a n x} d x = \frac{p + q}{2}

p = \int \sqrt{\frac{s i n x}{c o s x}} + \sqrt{\frac{c o s x}{s i n x}} d x

p = \int \frac{s i n x + c o s x}{\sqrt{s i n x c o s x}} d x

p = \sqrt{2} \int \frac{d (s i n x - c o s x)}{\sqrt{1 - 1 + 2 s i n x c o s x}} d x

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

= \sqrt{2} {s i n}^{- 1} (s i n x - c o s x) + c_{1}

q = \int (\sqrt{\frac{s i n x}{c o s x}} - \sqrt{\frac{c o s x}{s i n x}}) d x

q = \int \frac{s i n x - c o s x}{\sqrt{s i n x c o s x}} d x \to - \int \frac{c o s x - s i n x}{\sqrt{s i n x c o s x}} d x

q = (- \sqrt{2}) \int \frac{d (s i n x + c o s x)}{\sqrt{1 + 2 s i n x c o s x - 1}}

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

= (- \sqrt{2}) l n {(s i n x + c o s x) + \sqrt{{(s i n x + c o s x)}^{2} - 1}}

s o \int \sqrt{t a n x} d x

\frac{\sqrt{2}}{2} [{s i n}^{- 1} (s i n x - c o s x) - l n {(s i n x + c o s x) + \sqrt{{(s i n x + c o s x)}^{2} - 1}}] + c

p l s c h e c k \dots

Answered by MJS last updated on 26/Feb/19

∫(√(tan x))dx= [t=(√(tan x)) → dx=2cos^2 x (√(tan x))dt=((2tdt)/(t^4 +1))] =2∫(t^2 /(t^4 +1))dt=2∫(t^2 /((t^2 −(√2)t+1)(t^2 +(√2)t+1)))dt= =2∫((((√2)t)/(4(t^2 −(√2)t+1)))−(((√2)t)/(4(t^2 +(√2)t+1))))dt= =(1/2)∫((((√2)t−1)/(t^2 −(√2)t+1))+(1/(t^2 −(√2)t+1))−(((√2)t+1)/(t^2 +(√2)t+1))+(1/(t^2 +(√2)t+1)))dt= =((√2)/4)∫(((2t−(√2))/(t^2 −(√2)t+1))−((2t+(√2))/(t^2 +(√2)t+1)))dt+(1/2)∫((1/(t^2 −(√2)t+1))+(1/(t^2 +(√2)t+1)))dt= =((√2)/4)(ln (t^2 −(√2)t+1) −ln (t^2 +(√2)t+1))+((√2)/2)(arctan ((√2)t−1) +arctan ((√2)t+1))= =((√2)/4)(ln ∣((tan x −(√(2tan x))+1)/(tan x +(√(2tan x))+1))∣ +2(arctan ((√(2tan x))−1) +arctan ((√(2tan x))+1)))+C

\int \sqrt{\tan x} d x =

[t = \sqrt{\tan x} \to d x = {2 \cos}^{2} x \sqrt{\tan x} d t = \frac{2 t d t}{t^{4} + 1}]

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

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

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

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

= \frac{\sqrt{2}}{4} (\ln (t^{2} - \sqrt{2} t + 1) - \ln (t^{2} + \sqrt{2} t + 1)) + \frac{\sqrt{2}}{2} (\arctan (\sqrt{2} t - 1) + \arctan (\sqrt{2} t + 1)) =

= \frac{\sqrt{2}}{4} (\ln ∣ \frac{\tan x - \sqrt{2 \tan x} + 1}{\tan x + \sqrt{2 \tan x} + 1} ∣ + 2 (\arctan (\sqrt{2 \tan x} - 1) + \arctan (\sqrt{2 \tan x} + 1))) + C

Commented by behi83417@gmail.com last updated on 26/Feb/19

arctg((√(2tgx))−1)+arctg((√(2tgx))+1)= arctg((2(√(2tgx)))/(2(1−tgx)))=arctg(((√(2tgx))/(1−tgx)))= arccos∣cosx−sinx∣ ln((tgx−(√(2tgx))+1)/(tgx+(√(2tgx))+1))=ln(((tgx+1−(√(2tgx)))^2 )/(tg^2 x+1))= =2ln[cosx.(tgx+1−(√(2tgx))]= =2ln[sinx+cosx−(√(sin2x))] ⇒I=((√2)/2).ln∣cosx+sinx−(√(sin2x))∣+2cos^(−1) ∣cosx−sinx∣+C

a r c t g (\sqrt{2 t g x} - 1) + a r c t g (\sqrt{2 t g x} + 1) =

a r c t g \frac{2 \sqrt{2 t g x}}{2 (1 - t g x)} = a r c t g (\frac{\sqrt{2 t g x}}{1 - t g x}) =

a r c \cos ∣ c o s x - s i n x ∣

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

= 2 l n [c o s x . (t g x + 1 - \sqrt{2 t g x}] =

= 2 l n [s i n x + c o s x - \sqrt{s i n 2 x}]

\Rightarrow I = \frac{\sqrt{2}}{2} . \ln ∣ cosx + sinx - \sqrt{\sin 2 x} ∣ + 2 \cos^{- 1} ∣ cosx - sinx ∣ + C

Commented by MJS last updated on 26/Feb/19

thank you!

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