What is $\int {csc}^{4} (x) {cot}^{6} (x) d x$ $int csc^4(x) cot^6(x) dx$ ?

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What is $\int {csc}^{4} (x) {cot}^{6} (x) d x$ $int csc^4(x) cot^6(x) dx$ ?

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Answer 1 · 2016-04-17T13:53:43

$- \frac{{cot}^{9} (x)}{9} - \frac{{cot}^{7} (x)}{7} + C$ $-cot^9(x)/9-cot^7(x)/7+C$

Explanation:

Notice that when we have $cot (x)$ $cot(x)$ and $csc (x)$ $csc(x)$ functions, we will often have derivatives embedded into an integral. Here, the most important derivative to remember is

$\frac{d}{d x} (cot (x)) = - {csc}^{2} (x)$ $d/dx(cot(x))=-csc^2(x)$

Thus, if we let $u = cot (x)$ $u=cot(x)$ , then $\frac{d u}{d x} = - {csc}^{2} (x)$ $(du)/dx=-csc^2(x)$ and $d u = - {csc}^{2} (x) d x$ $du=-csc^2(x)dx$ .

However, we have $4$ $4$ different $csc (x)$ $csc(x)$ functions but we only want $2$ $2$ . We can turn the remaining ${csc}^{2} (x)$ $csc^2(x)$ term into functions of $cot (x)$ $cot(x)$ using the form of the Pythagorean identity:

$1 + {cot}^{2} (x) = {csc}^{2} (x)$ $1+cot^2(x)=csc^2(x)$

Thus, the integral becomes:

$\int {csc}^{4} (x) {cot}^{6} (x) d x = \int {csc}^{2} (x) {csc}^{2} (x) {cot}^{6} (x) d x$ $intcsc^4(x)cot^6(x)dx=intcsc^2(x)csc^2(x)cot^6(x)dx$

$= \int {csc}^{2} (x) [(1 + {cot}^{2} (x)) ({cot}^{6} (x)] d x$ $=intcsc^2(x)[(1+cot^2(x))(cot^6(x)]dx$

$= \int {csc}^{2} (x) [{cot}^{6} (x) + {cot}^{8} (x)] d x$ $=intcsc^2(x)[cot^6(x)+cot^8(x)]dx$

Now, in order to get a $d u = - {csc}^{2} (x) d x$ $du=-csc^2(x)dx$ term, we need to multiply the interior and exterior of the integral by $- 1$ $-1$ .

$= - \int (- {csc}^{2} (x)) [{cot}^{6} (x) + {cot}^{8} (x)] d x$ $=-int(-csc^2(x))[cot^6(x)+cot^8(x)]dx$

Now substitute, since we have out $d u$ $du$ term, recalling that $u = cot (x)$ $u=cot(x)$ :

$= - \int (u^{6} + u^{8}) d u$ $=-int(u^6+u^8)du$

Integrating term by term, this gives us

$= - (\frac{u^{7}}{7} + \frac{u^{9}}{9}) + C$ $=-(u^7/7+u^9/9)+C$

$= - \frac{{cot}^{9} (x)}{9} - \frac{{cot}^{7} (x)}{7} + C$ $=-cot^9(x)/9-cot^7(x)/7+C$

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What is $\int {csc}^{4} (x) {cot}^{6} (x) d x$ $int csc^4(x) cot^6(x) dx$ ?

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What is int csc^4(x) cot^6(x) dx ∫csc4(x)cot6(x)dxint csc^4(x) cot^6(x) dx ?

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What is $\int {csc}^{4} (x) {cot}^{6} (x) d x$ $int csc^4(x) cot^6(x) dx$ ?