How do you integrate ${csc}^{3} x$ $csc^3x$ ?

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Answer 1 · 2016-08-19T16:35:48

$\frac{- cot x csc x - ln (| cot x + csc x |)}{2} + C$ $(-cotxcscx-ln(abs(cotx+cscx)))/2+C$

We have:

$I = \int {csc}^{3} x d x$ $I=intcsc^3xdx$

We will use integration by parts. First, rewrite the integral as:

$I = \int {csc}^{2} x csc x d x$ $I=intcsc^2xcscxdx$

Since integration by parts takes the form $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ , let:

${(u=cscx" "=>" "du=-cotxcscxdx),(dv=csc^2xdx" "=>" "v=-cotx):}$

Applying integration by parts:

$I=-cotxcscx-intcot^2xcscxdx$

Through the Pythagorean identity, write $cot^2x$ as $csc^2x-1$ .

$I=-cotxcscx-int(csc^2x-1)(cscx)dx$

$I=-cotxcscx-intcsc^3xdx+intcscxdx$

Note that $I=intcsc^3xdx$ and $intcscxdx=-ln(abs(cotx+cscx))$ .

$I=-cotxcscx-I-ln(abs(cotx+cscx))$

Add the original integral $I$ to both sides.

$2I=-cotxcscx-ln(abs(cotx+cscx))$

Solve for $I$ and add the constant of integration:

$I=(-cotxcscx-ln(abs(cotx+cscx)))/2+C$

How do you integrate csc^3xcsc3x csc^3x?