Question

Как решать? можно с подробным решениям. интеграл сosx/tg(^5)x

Answer 1

$\int \frac{cosx\cdot dx}{tg^5x} =\int \frac{cosx\cdot dx}{ \frac{sin^5x}{cos^5x} } =\int \frac{cos^5x\cdot cosx\cdot dx}{sin^5x} =\\\\=[\, u=cos^5x,\; du=-5cos^4x\cdot sinx\, dx,\; dv= \frac{cosx\cdot dx}{sin^5x} ,\\\\v=\int \frac{cosx\, dx}{sin^5x}=\int \frac{d(sinx)}{sin^5x} =\int (sinx)^{-5}\cdot d(sinx)=\int t^{-5}dt=\\\\= \frac{t^{-4}}{-4}= \frac{(sinx)^{-4}}{-4} = -\frac{1}{4sin^4x} \; ;\; \; \int u\cdot dv=uv-\int v\, du\; ]=$

$=- \frac{cos^5x}{4sin^4x} -\int \frac{5cos^4x\cdot sinx\, dx}{4sin^4x} =- \frac{cos^5x}{4sin^4x} -\frac{5}{4}\, \int \frac{cos^4x\, dx}{sin^3x} =\\\\=[\, u=cos^3x,du=-3cos^2x\cdot sinx\, dx,dv= \frac{cosx\, dx}{sin^3x}=(sinx)^{-3}\cdot cosx\, dx ,$

$v=\int (sinx)^{-3}\cdot d(sinx)=\frac{(sinx)^{-2}}{-2}=- \frac{1}{2sin^2x}\; ]=\\\\=- \frac{cos^5x}{4sin^4x} - \frac{5}{4} \cdot (\frac{cos^3x}{2sin^2x} - \frac{3}{2}\, \int \frac{cos^2x\cdot sinx\, dx}{sin^2x} )=\\\\=- \frac{cos^5x}{4sin^4x} - \frac{5}{8} \cdot \frac{cos^3x}{sin^2x}+\frac{15}{8}\cdot \int \frac{cos^2x\, dx}{sinx} =\\\\=- \frac{cos^5x}{4sin^4x}-\frac{5}{8}\cdot \frac{cos^3x}{sin^2x} + \frac{15}{8} \cdot \int \frac{(1-sin^2x)dx}{sinx}=$

$= -\frac{cos^5x}{4sin^4x}-\frac{5cos^3x}{8sin^2x}+\frac{15}{8}\cdot (\int \frac{dx}{sinx}-\int sinx\, dx)=\\\\=[\, \int \frac{dx}{sinx}=(t=tg\frac{x}{2},\, sinx=\frac{2t}{1+t^2},dx= \frac{2dt}{1+t^2}\, )=\\\\=\int \frac{dt}{t}=ln|t|+C=ln|tg\frac{x}{2}|+C\, ]=$

$=-\frac{cos^5x}{4sin^4x}- \frac{5cos^3x}{8sin^2x} + \frac{15}{8}\cdot ln\left |tg\frac{x}{2}\right |+\frac{15}{8} \cdot cosx+C$