Question

помогитее пожалуйста​

Answer 1

Ответ:

6.

$\int\limits( \sin(x) + 5 \cos(x) )dx = - \cos(x) + 5\sin(x) + C$

7.

$\int\limits( \sin( \frac{x}{2} ) + \cos( \frac{x}{2} ) ) {}^{2} dx = \\ = \int\limits( { \sin }^{2} ( \frac{x}{2} ) + 2 \sin( \frac{x}{2} ) \cos( \frac{x}{2} ) + \cos {}^{2} ( \frac{x}{2} ) )dx = \\ = \int\limits(1 + \sin(x)) dx = x - \cos(x) + C$

8.

$\int\limits \frac{ \cos(2x) }{ \cos {}^{2} (x) \sin {}^{2} (x) } dx = \int\limits \frac{ \cos {}^{2} (x) - \sin {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } dx = \\ = \int\limits( \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } - \frac{ \sin {}^{2} (x) }{ \sin {}^{2} (x) \cos {}^{2} (x) } )dx = \\ = \int\limits( \frac{1}{ \sin {}^{2} (x) } - \frac{1}{ \cos {}^{2} (x) } )dx = - ctgx - tgx + C$

9.

$\int\limits \frac{ {x}^{4} }{1 + {x}^{2} } dx = \int\limits \frac{ {x}^{4} - 1 + 1 }{ {x}^{2} + 1} dx = \\ = \int\limits \frac{ {x}^{4} - 1 }{ {x}^{2} + 1} dx + \int\limits\frac{1}{1 + {x}^{2} } dx = \\ = \int\limits \frac{( {x}^{2} - 1)( {x}^{2} + 1) }{ {x}^{2} + 1} + arctgx = \\ = \int\limits( {x}^{2} - 1)dx + arctgx = \frac{ {x}^{3} }{3} - x + arctgx + C$

10.

$\int\limits \frac{3 - 2 {ctg}^{2}x }{ \cos {}^{2} (x) } dx = \int\limits( \frac{3}{ \cos {}^{2} (x) } - 2 \times \frac{1}{ \cos {}^{2} (x) } \times \frac{ \cos {}^{2} (x) }{ \sin {}^{2} (x) } )dx = \\ = \int\limits( \frac{3}{ \cos {}^{2} (x) } - \frac{2}{ \sin {}^{2} (x) } )dx = 3tgx + 2ctgx + C$

11.

$\int\limits \frac{1 - \sin {}^{3} (x) }{ \sin {}^{2} (x) } dx= \int\limits( \frac{1}{ \sin {}^{2} (x) } - \frac{ \sin {}^{3} (x) }{ \sin {}^{2} ( {}^{}x ) } )dx = \\ = - ctgx -\int\limits \sin(x) dx = - ctgx + \cos(x) + C$

Answer 2

Ответ:

$6.\ \ \int (sinx+5cosx)\, dx=\int sinx\, dx-5\int cosx\, dx=-cosx-5\, sinx+C\\\\\\7.\ \ \int \Big(sin\dfrac{x}{2}+cos\dfrac{x}{2}\Big)^2\, dx=\int \Big(sin^2\dfrac{x}{2}+2\, sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}+cos^2\dfrac{x}{2}\Big)\, dx=\\\\\\=\int (1+sinx)\, dx=x-cosx+C\\\\\\8.\ \ \int \dfrac{cos2x}{cos^2x\cdot sin^2x}\, dx=\int \dfrac{cos^2x-sin^2x}{cos^2x\cdot sin^2x}\, dx=\int \dfrac{cos^2x}{sin^2x\cdot cos^2x}\, dx-$

$-\int \dfrac{sin^2x}{cos^2x\cdot sin^2x}\, dx=\int \dfrac{dx}{sin^2x}-\int\dfrac{dx}{cos^2x}=-ctgx-tgx+C=\\\\\\=-(ctgx+tgx)+C$

$9.\ \ \int \dfrac{x^4}{1+x^2}\, dx=\int \Big(x^2-1+\dfrac{1}{1+x^2}\Big)\, dx=\dfrac{x^3}{3}-x+arctgx+C\\\\\\10.\ \ \int \dfrac{3-2ctg^2x}{cos^2x}\, dx=\int \Big(\dfrac{3}{cos^2x}-\dfrac{2}{sin^2x}\Big)\, dx=3\, tgx+2\, ctgx+C\\\\\\11.\ \ \int \dfrac{1-sin^3x}{sin^2x}\, dx=\int \Big(\dfrac{1}{sin^2x}-sinx\Big)\, dx=-ctgx+cosx+C$