Question

решить интеграл тригонометрической функции: $\int\limits {\frac{cos^4x}{sin^3x} } \, dx$ . ответ должен получиться: $- \frac{3}{2} cos x - \frac{cos^3x}{2sin^2x} - \frac{3}{2} ln|tg\frac{x}{2} | + C$

Answer 1

$\displaystyle \int \frac{cos^4x}{sin^3x}\, dx=\int \frac{(cos^2x)^2}{sin^3x}\, dx=\int \frac{(1-sin^2x)^2}{sin^3x}\, dx=\int \frac{1-2sin^2x+sin^4x}{sin^3x}\, dx=\\\\\\=\int \Big(\frac{1}{sin^3x}-\frac{2}{sinx}+sinx\Big)\, dx=\frac{1}{2}\, ln\Big|tg\frac{x}{2}\Big|-\frac{cosx}{2sin^2x} -2\, ln\Big|tg\frac{x}{2}\Big|-cosx+C=\\\\\\=-\frac{3}{2}\, ln\Big|tg\frac{x}{2}\Big|-\frac{cosx}{2sin^2x}-sinx+C\ ;$

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

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