Question

Помогите решить интеграл: $\int\limits {\frac{dx}{sin^3x} }$ . Ответ должен выйти: $\frac{1}{2} ln|tg\frac{x}{2} |- \frac{cos x}{2sin^2x} + C$

Answer 1

Ответ:

$\displaystyle \int \frac{dx}{sin^3x}=\int \frac{sin^2x+cos^2x}{sin^3x}\, dx=\int \frac{dx}{sinx}+\int \frac{cos^2x}{sin^3x}\, dx=\\\\\\=\int \frac{dx}{2sin\frac{x}{2}\cdot cos\frac{x}{2}}+\int \underbrace {cosx}_{u}\cdot \underbrace {\frac{cosx\, dx}{sin^3x}}_{dv}=\int \frac{\dfrac{1}{2}\cdot \dfrac{dx}{cos^2\frac{x}{2}}}{\dfrac{sin\frac{x}{2}\cdot cos\frac{x}{2}}{cos^2\frac{x}{2}}}+\Big[\ u=cosx\ ,$

$\displaystyle du=-sinx\, dx\ ,\ dv=\frac{cosx\, dx}{sin^3x}\ ,\ v=\int (sinx)^{-3}\cdot d(sinx)=-\frac{1}{2\, sin^2x}\Big]=\\\\\\=\int \frac{d(tg\frac{x}{2})}{tg\frac{x}{2}}-\frac{cosx}{2sin^2x}-\int \frac{sinx\, dx}{2sin^2x}=ln\Big|tg\frac{x}{2}\Big|-\frac{cosx}{2sin^2x}-\frac{1}{2}\int \frac{dx}{sinx}=\\\\\\=ln\Big|tg\frac{x}{2}\Big|-\frac{cosx}{2sin^2x}-\frac{1}{2}\cdot ln\Big|tg\frac{x}{2}\Big|+C=\frac{1}{2}\cdot ln\Big|tg\frac{x}{2}\Big|-\frac{cosx}{2sin^2x} +C$