Question

Como resolver esta integral por partes. Ayuden por favor!!

a) ∫csc³(x)dx

Answer 1

recuerda que $(\cot x)'=-\csc^2x$ entonces hacemos lo siguiente

$\displaystyle \int \csc^3 x\, dx=-\int \csc x\cdot (-\csc^2x)\, dx\\ \\ \\ \int \csc^3 x\, dx=-\int \csc x\, d(\cot x)\\ \\ \\ \int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \cot x\cdot d(\csc x)\\ \\ \\ \int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \cot x (-\csc x \cot x)\,dx\\ \\ \\ \int \csc^3 x\, dx=-\csc x\cdot\cot x-\int\csc x\cot^2 x\, dx$

$\displaystyle \int \csc^3 x\, dx=-\csc x\cdot\cot x-\int\csc x\cdot (\csc^2x-1)\, dx\\ \\ \\ \int \csc^3 x\, dx=-\csc x\cdot\cot x-\int \csc^3x-\csc x \, dx\\ \\ \\ 2\int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \csc x \, dx\\ \\ \\ 2\int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \dfrac{1}{\sin x} \, dx\\ \\ \\ 2\int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \dfrac{\sin x}{\sin^2 x} \, dx$

$\displaystyle 2\int \csc^3 x\, dx=-\csc x\cdot\cot x+\int \dfrac{\sin x}{1-\cos^2 x} \, dx\\ \\ \\ 2\int \csc^3 x\, dx=-\csc x\cdot\cot x-\int \dfrac{d(\cos x)}{1-\cos^2 x} \\ \\ \\ \text{Recuerde que... } \int \dfrac{1}{1-u^2}\, du=\dfrac{1}{2}\ln \left|\dfrac{1+u}{1-u}\right|+C\\ \\ \\ 2\int \csc^3 x\, dx=-\csc x\cdot\cot x-\dfrac{1}{2}\ln \left|\dfrac{1+\cos x}{1-\cos x}\right|+C$


$\displaystyle 2\int \csc^3 x\, dx=-\csc x\cdot\cot x-\ln |\csc x + \cot x|+C\\ \\ \\ \int \csc^3 x\, dx=-\dfrac{1}{2}\csc x\cdot\cot x-\dfrac{1}{2}\ln |\csc x + \cot x|+C$