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)). Created by Муса134. matematika-ru

(1+t^2))...

NNNLLL54

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

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

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oganesbagoyan

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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) .

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