Question

найти интеграл dx/(3sinx+4cosx) (замена t=tg(x/2),sinx=(2t)/(1+t^2),dx=(2dt)/(1+t^2),cosx=(1-t^2)/(1+t^2))​

Answer 1

$int dfrac{dx}{3sinx+4cosx}=Big[; t=tgdfrac{x}{2},; sinx=dfrac{2t}{1+t^2},; cosx=dfrac{1-t^2}{1+t^2},; dx=dfrac{2, dt}{1+t^2}; Big]=\\\=int frac{2, dt}{frac{6t}{1+t^2}+frac{4-4t^2}{1+t^2}}=int dfrac{2(1+t^2)}{-2, (2t^2-3t-2)}, dt=-dfrac{1}{2}int , dfrac{2t^2+2}{2t^2-3t-2}, dt=\\\=-dfrac{1}{2}int Big(, 1+dfrac{3t+4}{(2t+1)(t-2)}Big), dt=Q$

$dfrac{3t+4}{(2t+1)(t-2)}=dfrac{A}{2t+1}+dfrac{B}{t-2}\\3t+4=A(t-2)+B(2t+1)\\t=2:; ; B=frac{10}{5}=2\\t=-frac{1}{2}:; ; A=frac{5}{-5}=-1\\\Q=-frac{1}{2}int Big(1+dfrac{-1}{2t+1}+dfrac{2}{t-2}Big), dt=\\=-frac{1}{2}cdot Big(t-frac{1}{2}, ln|2t+1|+2, ln|t-2|Big)+C=\\=-frac{1}{2}, tgfrac{x}{2}+frac{1}{4}, ln|2tgfrac{x}{2}+1|-frac{1}{2}, ln|tgfrac{x}{2}-2|+C$

Answer 2

https://znanija.com/task/34692164

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Найти ∫ dx / (3sinx+4cosx)

Ответ:   ( Ln | C(2tg(x/2)+1 ) /( tg(x/2) - 2 ) |  ) / 5

Пошаговое объяснение:  

sinx = 2t /(1+t²)

cosx = (1 - t²)/(1+t²)

dx = (2dt) / (1+t²).

∫ (2dt) /(1+t²)  / ( 6t / (1+t²) +4(1-t²) / (1+t²) )    =

= ∫ (dt) / ( 3t  +2(1-t²)  )    =  - ∫ (dt) / ( 2t² - 3t  -2 )  = - ∫ (dt) / ( t-2)(2t +1 )=

(1/5) *∫( 2/(2t+1) - 1/(t - 2) ) dt =(1/5)*[ ∫( (2dt) / (2t +1)  - ∫(dt ) /( t-2) ] =

(1/5)* ( Ln|2t+1| - Ln|t-2| +Ln|C| )= (1/5)*Ln |C(2t+1) / (t-2) |  =

( Ln | C(2tg(x/2)+1 ) /( tg(x/2) - 2 ) |  ) / 5 .   

Универсальная  замена  :   t =   tg(x/2)               ⇒      

dt =(1 / cos²(x/2) ) *(1/2) dx  =( 1+tg²(x/2) )*(1/2)*dx   dx = (2dt) / (1+t²)

sinx =  2sin(x/2)*cos(x/2)  = 2sin(x/2)*cos(x/2) / ( sin²(x/2) + cos²(x/2) )  =  2tg(x/2) / ( 1+tg²(x/2 )   = 2t /(1+t²)

cosx = (cos²(x/2) -sin²(x/2) )/( cos²(x/2) +sin²(x/2) ) =(1-tg²(x/2) )/(1+tg²(x/2)) = (1 - t²)/(1+t²) .