How do you integrate #cscxcot^4xdx#

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Answer 1 · 2015-05-04T16:55:13

#intcsc(x)*cot^4(x)dx#

Just trigonometric function : let's #t = tan(1/2x)#

#csc(x) = (1+t^2)/(2t)#

#cot(x) = (1-t^2)/(2t)#

#dx = 2/(1+t^2)dt#

So now we have :

#int(1+t^2)/(2t)*((1-t^2)/(2t))^4*2/(1+t^2)dt#

#=> int 1/t*((1-t^2)/(2t))^4dt#

#=>int1/t*(1-4t^2+6t^4-4t^6+t^8)/(16t^4)dt#

#=>1/16int(1-4t^2+6t^4-4t^6+t^8)/(t^5)dt#

#=>1/16int1/t^5-4/t^3+6/t-4t+t^3dt#

#=>1/16[-1/(4t^4)+2/t^2+6ln(|t|)-2t^2+1/4t^4]+C#

And then substitute back for #t = tan(1/2x)#...