Question

1)
$\int\limits {\frac{1}{3+4sinx} } \, dx$

2)
$\int\limits {\frac{sin^3x}{cosx-3} } \, dx$

Answer 1

Ответ: во вложении Объяснение:

Answer 2

$1)\; \; \int \frac{dx}{3+4sinx}=\Big [\; t=tg\frac{x}{2}\; ,\; sinx=\frac{2t}{1+t^2}\; ,\; dx=\frac{2\; dt}{1+t^2}\; \Big ]=\int \frac{2\; dt}{(1+t^2)\cdot (3+\frac{4\cdot 2t}{1+t^2})}=\\\\=2\int \frac{dt}{3t^2+8t+3}=2\int \frac{dt}{3\; (t^2+\frac{8}{3}\, t+1)}=\frac{2}{3}\int \frac{d(t+\frac{4}{3})}{(t+\frac{4}{3})^2-\frac{7}{9}}=$

$=\frac{2}{3}\cdot \frac{1}{2\cdot \frac{\sqrt7}{3}}\cdot ln\Big |\frac{t+\frac{4}{3}-\frac{\sqrt7}{3}}{t+\frac{4}{3}+\frac{\sqrt7}{3}}\Big |+C=\frac{1}{\sqrt{7}}\cdot ln\Big |\frac{3tg\frac{x}{2}+4-\sqrt{7}}{3tg\frac{x}{2}+4+\sqrt{7}}\Big |+C$

$2)\; \; \int \frac{sin^3x}{cosx-3}\, dx=\int \frac{sin^2x\cdot sinx\, dx}{cosx-3}=\int \frac{(1-cos^2x)\cdot sinx\, dx}{cosx-3}=\\\\=\Big [\; t=cosx\; ,\; dt=-sinx\, dx\; \Big ]=-\int \frac{(1-t^2)\cdot dt}{t-3}=\int \frac{t^2-1}{t-3}\, dt=\\\\=\int (t+3+\frac{8}{t-3})\, dt=\frac{t^2}{2}+3t+8\, ln|t-3|+C=\\\\=\frac{cos^2x}{2}+3\, cosx+8\, ln|cosx-3|+C$