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# 2 ways

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629
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Are both ways correct ?  Apr 15, 2015

#4
+5

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.

Apr 16, 2015

#1
+5

Hi  315 I do not know WHAT you are doing BUT the second one is NOT correct.

This is how I would do it.

$$\\\int\;CotxCosec^4x\;dx\\\\ =\int\;\frac{Cosx}{Sinx}\frac{1}{sin^4x}\;dx\\\\ =\int\;\frac{Cosx}{Sin^5x}\;dx\\\\ =\int\;CosxSin^{-5}x\;dx\\\\ =\frac{Sin^{-4}x}{-4}+c\\\\ =\frac{-Cosec^{4}x}{4}+c\\\\$$

.
Apr 15, 2015
#2
+5

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.

Apr 15, 2015
#3
+5

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)

Apr 15, 2015
#4
+5

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.

Bertie Apr 16, 2015