Решить интеграл [tex]\int\limits^a_b \frac{2-sinx}{sin^{2} x} dx[/tex]. Created by Rixo. matematika-ru

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Решить интеграл
[tex]\int\limits^a_b \frac{2-sinx}{sin^{2} x} dx[/tex]

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Alexаndr

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$\displaystyle \int\limits^a_b\frac{2-sinx}{sin^2x}dx=\int\limits^{tg\frac{a}{2}}_{tg\frac{b}{2}}(\frac{(1+t^2)^2}{4t^2}*\frac{2t^2-2t+2}{1+t^2}*\frac{2}{1+t^2})dt=\\=\int\limits^{tg\frac{a}{2}}_{tg\frac{b}{2}}\frac{t^2-t+1}{t^2}dt=\int\limits^{tg\frac{a}{2}}_{tg\frac{b}{2}}(1-\frac{1}{t}+\frac{1}{t^2})=(t-ln|t|-\frac{1}{t})|^{tg\frac{a}{2}}_{tg\frac{b}{2}}=\\=tg\frac{a}{2}-tg\frac{b}{2}-ln|\frac{tg\frac{a}{2}}{tg\frac{b}{2}}|-(\frac{1}{tg\frac{a}{2}}-\frac{1}{tg\frac{b}{2}})$

$\displaystyle t=tg\frac{x}{2}\to 2arctgt=x\to dx=\frac{2dt}{1+t^2}\\sinx=\frac{2t}{1+t^2}\\2-sinx=2-\frac{2t}{1+t^2}=\frac{2t^2-2t+2}{1+t^2}\\sin^2x=\frac{4t^2}{(1+t^2)^2}\to\frac{1}{sin^2x}=\frac{(1+t^2)^2}{4t^2}$

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