

 бах8 
                                                26.12.2020 00:29
                                                
                                                бах8 
                                                26.12.2020 00:29
                                            ![\int \dfrac{dx}{sinx+tgx}=\int \dfrac{dx}{sinx+\frac{sinx}{cosx}}=\int \dfrac{cosx\, dx}{sinx\cdot cosx+sinx}=\int \dfrac{cosx\, dx}{sinx\cdot (cosx+1)}=\\\\\\=\Big[\ t=tg\dfrac{x}{2}\ ,\ sinx=\dfrac{2t}{1+t^2}\ ,\ cosx=\dfrac{1-t^2}{1+t^2}\ ,\ dx=\dfrac{2\, dt}{1+t^2}\Big]=\\\\\\=\int \dfrac{\dfrac{1-t^2}{1+t^2}\cdot \dfrac{2\, dt}{1+t^2}}{\dfrac{2t}{1+t^2}\cdot \Big(\dfrac{1-t^2}{1+t^2}+1\Big)}=\int \dfrac{2(1-t^2)\, dt}{2t\cdot (1-t^2+1+t^2)}=\int \dfrac{(1-t^2)\, dt}{2t}=](/tpl/images/1509/8453/9f97d.png)

 
                                                 p1pal1 
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                                                p1pal1 
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