Question

Срочно!!

Помогите пожалуйста
В первом это cosec

Answer 1

Ответ:

Объяснение:

1)  y = csec(4x - 1 );   y' = - ctg(4x -1 )csec( 4x- 1 )* ( 4x - 1 )' =

= - 4 ctg(4x -1 )csec( 4x- 1 ) ;

2) y = x²/( 1 + cos4x ) ;    y'= [ 2x( 1 + cos4x ) -x²( - 4sin4x ) ]/( 1 + cos4x )² =

= [ 2x( 1 + cos4x + 2xsin4x) ]/( 1 + cos4x )² ;

3)  ∫ dx/( 1 + 2x²) = 1/2 ∫ dx/(x² + (√2 /2)²) = 1/2 * 2/√2 *arctgx/(√2 /2) + C =

= √2 /2 arctgx/√2  + C ;

4)    ∫ 2xdx/√( 3x ) = 2/√3 ∫ xdx/x^(1/2) = 2/√3 ∫ x^(1/2 )dx = 2/√3 * 1/2 X      

X x^( -1/2)  + C = 1/√3 *1/√x + C = 1/√( 3x ) + C .

Answer 2

Ответ:

$1)\ \ y=cosec(4x-1)=\dfrac{1}{sin(4x-1)}\ \ ,\ \ \ \ \Big(\dfrac{C}{v}\Big)'=-\dfrac{C\, v'}{v^2} \\\\\\y'=-\dfrac{4cos(4x-1)}{sin^2(4x-1)}\\\\\\2)\ \ y=\dfrac{x^2}{1+cos4x}\\\\\\y'=\dfrac{2x\, (1+cos4x)-x^2\cdot (-4sin4x)}{(1+cos4x)^2}=\dfrac{2x+2x\, cos4x+4x^2\, sin4x}{(1+cos4x)^2}$

$3)\ \ \displaystyle \int \frac{dx}{1+2x^2}=\int \frac{dx}{1+(\sqrt2x)^2}=\frac{1}{\sqrt2}\, arctg(\sqrt2x)+C\\\\\\4)\ \ \int \frac{2x\, dx}{\sqrt{3x}}=\frac{2}{\sqrt3}\int \frac{x\, dx}{\sqrt{x}}=\frac{2}{\sqrt3}\int \sqrt{x}\, dx=\frac{2}{\sqrt3}\cdot \frac{2\sqrt{x^3}}{3}+C=\frac{4\sqrt{x^3}}{3\sqrt3}+C$