Зайки, умоляю. не проходите мимо!!ОЧЕНЬ НУЖНО. Решить методом замены переменной. что...

$17) \int{\sin^3(x)\ dx} = -\int{\sin^2(x)\ d\cos(x)} =$

$\sin^2(x)=1 - \cos^2(x)$

$= -\int{(1 - \cos^2(x))\ d\cos(x)} = -\int{d\cos(x)} + \int{\cos^2(x)\ d\cos(x)} =$

$= \cos(x) + \frac{cos^3(x)}{3} + C$

$18) \int{\frac{1}{x^2 - 9}}dx = \int{\frac{1}{x^2 - 3^2}}dx = \frac{1}{6}\ln|\frac{x-3}{x+3}| + C$

$20) \int{\frac{x^2}{x^2 + 2}}dx = \int{(1 - \frac{2}{x^2 + 2})}dx = \int{dx} - 2\cdot \int{\frac{dx}{x^2 + 2}} =$

$= x - 2\cdot \frac{1}{\sqrt{2}}\arctan(\frac{x}{\sqrt{2}}) + C$

$22) \int{\frac{1}{\sqrt{4-9x^2}}}dx = \int{\frac{1}{\sqrt{2^2-(3x)^2}}}dx = \arcsin(\frac{3x}{2}) + C$

$24) \int{x^2 \sin(3x^3)}dx = \frac{1}{3}\int{\sin(3x^3)}d(x^3) = \frac{1}{9}\int{\sin(3x^3)}d(3x^3) =$

$=-\frac{1}{9}\cos(3x^3) + C$