Question

Помогите решить интегралы.

Answer 1

Пошаговое объяснение:

$\int {\frac{x \sin 2x}{\sin ^3x} } \, dx =\begin{Vmatrix} u=x&du=dx\\dv=\frac{\sin 2x}{\sin ^3x}dx &v=-\frac{2}{\sin x} \\\int {\frac{\sin 2x}{\sin ^3x} } \, dx =\int {\frac{2 \sin x \cos x}{\sin ^3x} } \, dx &=2\int {\frac{d(\sin x) }{\sin ^2x} } \, =-\frac{2}{\sin x} +C\end{Vmatrix}=$

$-\frac{2x}{\sin x} -\int {-\frac{2}{\sin x} } \, dx =-\frac{2x}{\sin x} +2\int {\frac{ \sin x}{\sin ^2x} } \, dx=\begin{Vmatrix}d( \cos x)= -\sin xdx\end{Vmatrix}=$

$-\frac{2x}{\sin x} -2\int {\frac{d( \cos x)}{1 - \cos ^2x} } \,=-\frac{2x}{\sin x} -2*\frac{1}{2} \ln|\frac{1+ \cos x}{1 - \cos x} |+C=-(\frac{2x}{\sin x}+ \ln|\frac{1+ \cos x}{1 - \cos x} |)+C$

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$\int {\frac{dx}{5+3 \cos x} } \, =\begin{Vmatrix} t= \tan \frac{x}{2} \\\cos x = \frac{1-t^2}{1+t^2} \\dx=\frac{2}{1+t^2}dt \end{Vmatrix}=\int {\frac{\frac{2}{1+t^2} }{5+3\frac{1-t^2}{1+t^2} } } \,dt =\int {\frac{\frac{2}{1+t^2} }{\frac{5+5t^2+3-3t^2}{1+t^2} } } \,dt=$

$\int {\frac{\frac{2}{1+t^2} }{\frac{8+2t^2}{1+t^2} } } \,dt=\int {\frac{2(1+t^2)}{2(1+t^2)(4+t^2)} } \, dt=\int {\frac{1}{(2^2+t^2)} } \, dt=\frac{1}{2} \arctan\frac{t}{2}+C= \frac{1}{2} \arctan(\frac{\tan \frac{x}{2} }{2})+C$