What is #int csc^4(x) cot^6(x) dx #?

Question

What is #int    csc^4(x) cot^6(x) dx #?

Charles Arocha · Accepted Answer

#-cot^9(x)/9-cot^7(x)/7+C#

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

#d/dx(cot(x))=-csc^2(x)#

Thus, if we let #u=cot(x)#, then #(du)/dx=-csc^2(x)# and #du=-csc^2(x)dx#.

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

#1+cot^2(x)=csc^2(x)#

As a result, the integral becomes:

#intcsc^4(x)cot^6(x)dx=intcsc^2(x)csc^2(x)cot^6(x)dx#

#=intcsc^2(x)[(1+cot^2(x))(cot^6(x)]dx#

#=intcsc^2(x)[cot^6(x)+cot^8(x)]dx#

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

#=-int(-csc^2(x))[cot^6(x)+cot^8(x)]dx#

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

#=-int(u^6+u^8)du#

Adding term by term, this provides us with

#=-(u^7/7+u^9/9)+C#

#=-cot^9(x)/9-cot^7(x)/7+C#

Scarlett Allison · Answer

undefined To integrate $\int \csc^4(x) \cot^6(x) \, dx$, use the trigonometric identity $\csc^2(x) = \cot^2(x) + 1$.

First, rewrite the integral using this identity:
$$
\int \csc^4(x) \cot^6(x) \, dx = \int \csc^2(x) (\cot^2(x) \cot^4(x)) \, dx
$$

Substitute $\csc^2(x) = \cot^2(x) + 1$:
$$
\int (\cot^2(x) + 1) (\cot^2(x) \cot^4(x)) \, dx = \int (\cot^6(x) + \cot^2(x) \cot^4(x)) \, dx
$$

Now, let's integrate each term separately:

1) $\int \cot^6(x) \, dx$:

Rewrite $\cot^6(x)$ as $\cot^2(x) \cdot \cot^4(x)$:
$$
\int \cot^6(x) \, dx = \int \cot^2(x) \cdot \cot^4(x) \, dx
$$

Let $u = \cot(x)$ and $du = -\csc^2(x) \, dx$.
The integral becomes:
$$
-\int u^4 \, du = -\frac{1}{5}u^5 + C = -\frac{1}{5}\cot^5(x) + C
$$

2) $\int \cot^2(x) \cot^4(x) \, dx$:

Let $w = \cot(x)$ and $dw = -\csc^2(x) \, dx$.
The integral becomes:
$$
-\int w^3 \, dw = -\frac{1}{4}w^4 + C = -\frac{1}{4}\cot^4(x) + C
$$

Combine the results:
$$
\int \csc^4(x) \cot^6(x) \, dx = -\frac{1}{5}\cot^5(x) - \frac{1}{4}\cot^4(x) + C
$$