Ответ:
Найти неопределённый интеграл .
[tex]\bf \displaystyle 7)\int \frac{cos2x}{cosx-sinx}\, dx=\\\\\\=\int \frac{cos^2x-sun^2x}{cosx-sinx}\, dx=\int \frac{(cosx-sinx)(cosx+sinx)}{cosx-sinx}\,dx=\\\\\\=\int (cosx+sinx)\, dx=sinx-cosx+C\\\\\\8)\ \ \int \Big(5^{x}-\frac{3}{cos^2x}+\frac{4}{\sqrt{1-x^2}}\Big)\, dx=\int \,5, ^{x}\, dx-\int \frac{3\, dx}{cos^2x}+\int \frac{4\, dx}{\sqrt{1-x^2}}=\\\\\\=\frac{5^{x}}{ln\, 5}-3\cdot tg\, x+4\cdot arcsin\, x+C[/tex]
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Ответ:
Найти неопределённый интеграл .
[tex]\bf \displaystyle 7)\int \frac{cos2x}{cosx-sinx}\, dx=\\\\\\=\int \frac{cos^2x-sun^2x}{cosx-sinx}\, dx=\int \frac{(cosx-sinx)(cosx+sinx)}{cosx-sinx}\,dx=\\\\\\=\int (cosx+sinx)\, dx=sinx-cosx+C\\\\\\8)\ \ \int \Big(5^{x}-\frac{3}{cos^2x}+\frac{4}{\sqrt{1-x^2}}\Big)\, dx=\int \,5, ^{x}\, dx-\int \frac{3\, dx}{cos^2x}+\int \frac{4\, dx}{\sqrt{1-x^2}}=\\\\\\=\frac{5^{x}}{ln\, 5}-3\cdot tg\, x+4\cdot arcsin\, x+C[/tex]