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))​

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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) .