+893 You already have the method Melody, you used it in your earlier post.
To illustrate what's going on here, consider the, integrand cos(x).sin^3(x).
We can integrate that in, at least, two different ways.
We can integrate it directly as sin^4(x)/4 or we can use the trig identity sin^2 = 1 - cos^2 to get
cos(x)(1-cos^2(x))sin(x) = cos(x)sin(x) - cos^3(x).sin(x) which integrates to -cos^2(x)/2 + cos^4(x)/4.
The problem we are discussing is dealt with in exactly the same way except that the functions are cosec and cot.
Come back if you need further details.
+118608 Hi 315 
I do not know WHAT you are doing BUT the second one is NOT correct.
This is how I would do it.
$$\\\int\;CotxCosec^4x\;dx\\\\
=\int\;\frac{Cosx}{Sinx}\frac{1}{sin^4x}\;dx\\\\
=\int\;\frac{Cosx}{Sin^5x}\;dx\\\\
=\int\;CosxSin^{-5}x\;dx\\\\
=\frac{Sin^{-4}x}{-4}+c\\\\
=\frac{-Cosec^{4}x}{4}+c\\\\$$
+893 Both are correct.
Simply make use of the identity 1+cot^2 = cosec^2 to go from the first form to the second. The -1/4 gets lumped in with the constant of integration.
+118608 Thanks Bertie,
Sorry 315,
Bertie or 315, can you try to explain to me the method that you have used?
(Just the first one for starters)
+893 You already have the method Melody, you used it in your earlier post.
To illustrate what's going on here, consider the, integrand cos(x).sin^3(x).
We can integrate that in, at least, two different ways.
We can integrate it directly as sin^4(x)/4 or we can use the trig identity sin^2 = 1 - cos^2 to get
cos(x)(1-cos^2(x))sin(x) = cos(x)sin(x) - cos^3(x).sin(x) which integrates to -cos^2(x)/2 + cos^4(x)/4.
The problem we are discussing is dealt with in exactly the same way except that the functions are cosec and cot.
Come back if you need further details.
