Question

Помогите пожалуйста с решением , интеграл ​

Answer 1

ответы прикрепил на фотографии

Answer 2

Объяснение:

$\displaystyle\\15.\ \int\frac{dx}{3sinx+4cosx-3} .$

          Упростим выражение: 3sinx+4cosx-3

                                     Подстановка

 $\displaystyle\\u=tg\frac{x}{2} .\ \ \ \ \ \ \Rightarrow\ \ \ \ \ \ du=(tg\frac{x}{2})'=\frac{1}{cos^2\frac{x}{2} } *(\frac{x}{2})'=\frac{dx}{2cos^2\frac{x}{2} }\\\\dx=2cos^2\frac{x}{2} du \\\\1+u^2=1+tg^2\frac{x}{2} =1+\frac{sin^2\frac{x}{2} }{cos^2x}=\frac{cos^2\frac{x}{2} +sin^2\frac{x}{2} }{cos^2\frac{x}{2} } =\frac{1}{cos^2\frac{x}{2} } \ \ \ \ \ \ \Rightarrow\\\\dx=\frac{2}{1+u^2} du.$

$\displaystyle\\1)\ sinx=2sin\frac{x}{2} cos\frac{x}{2} =2*\frac{sin\frac{x}{2} *cos\frac{x}{2} *cos\frac{x}{2} }{cos\frac{x}{2} } =2*tg\frac{x}{2} *cos^2\frac{x}{2}=\frac{2tg\frac{x}{2} }{\frac{1}{cos^2\frac{x}{2} } } =\\\\=\frac{2tg\frac{x}{2} }{\frac{sin^2\frac{x}{2}+cos^2\frac{x}{2} }{cos^2\frac{x}{2} } } =\frac{2tg\frac{x}{2} }{tg^2\frac{x}{2}+1 } =\frac{2u}{u^2+1} .$

$\displaystyle\\2)\ cosx=cos^2\frac{x}{2} -sin^2x=\frac{\frac{cos^2\frac{x}{2}-sin^2\frac{x}{2} }{cos^2x} }{\frac{1}{cos^2\frac{x}{2} } }=\frac{1-tg^2\frac{x}{2} }{tg^2\frac{x}{2}+1 } =\frac{1-u^2}{1+u^2} .$

$\displaystyle\\3)\ 3*\frac{2u}{1+u^2} +4*\frac{1-u^2}{1+u^2}-3=\frac{6u+4*(1-u^2)-3*(1+u^2)}{1+u^2} =\\\\=\frac{6u+4-4u^2-3-3u^2}{1+u^2} =\frac{-7u^2+6u+1}{1+u^2 }{cos^2\frac{x}{2} } } =-\frac{7u^2-6u-1}{1+u^2 } =\\\\=-\frac{7u^2-7u+u-1}{1+u^2} =-\frac{7u*(u-1)+(u-1)}{1+u^2}=-\frac{(u-1)(7u+1)}{1+u^2}\ \ \ \ \ \ \Rightarrow$

$\displaystyle\\\int(-\frac{(1+u^2)2}{(u-1)(7u+1)(1+u^2)} du=-\int\frac{2}{(u-1)(7u+1)}du.$

                                        Подстановка

$\displaystyle\\v=\frac{u-1}{7u+1} \ \ \ \ \Leftrightarrow\ \ \ \ u-1=(7u+1)*v\ \ \ \ \ \ \Leftrightarrow\ \ \ \ \ \ u-1=7uv+v\\\\7uv-u=-v-1\\\\u*(7v-1)=-(v+1)\\\\u=-\frac{v+1}{7v-1}\ |*\frac{7}{7} \\\\u=-\frac{7v+7}{7*(7v-1)} =-\frac{7v-1+8}{7*(7v-1)}=-\frac{8}{7*(7v-1)}-\frac{1}{7} \\\\du=(-\frac{8}{7*(7v-1)}-\frac{1}{7})'=-\frac{8'*(7*(7v-1)-8*(7*(7v-1))'}{7^2*(7v-1)^2}=\\\\=-\frac{0-8*7*7}{49*(7v-1)^2}dv=-\frac{-8}{(7v-1)^2}dv=\frac{8}{(7v-1)^2}dv \ \ \ \ \ \ \Rightarrow$

$\displaystyle\\u-1=-\frac{v+1}{7v-1} -1=\frac{-v-1-7v+1}{7v-1} =-\frac{8v}{7v-1} .\\\\7u+1=\frac{-7v-7+7v-1}{7v-1}=-\frac{8}{7v-1}. \ \ \ \ \ \ \Rightarrow\\\\-\int\frac{2}{(-\frac{8v}{7v-1})*(-\frac{8}{7v-1}) } *\frac{8}{(7v-1)^2}dv=-\int\frac{16*(7v-1)^2}{64v*(7v-1)^2} dv=\\\\=-\int\frac{dv}{4v} =-\frac{1}{4} *\int\frac{dv}{v}=-\frac{1}{4}*ln|v|+C =-\frac{1}{4}*ln|\frac{u-1}{7u+1}|+C=\\\\ =-\frac{1}{4}*(ln|u-1|-ln|7u+1|+C=\frac{ln|7tg\frac{x}{2}+1| }{4} -\frac{ln|tg\frac{x}{2}-1| }{4}+C .$