Here's how I would do it: csc(x)= 1/sin(x) so this is the same as ∫ dx/sin(x)- an odd power of x. Back in calculus, I learned that when you have an odd power of either sin(x) or cos(x), you can use one of them with dx and use sin2+ cos2= 1 for the rest: here sin(x) is in the denominator so I would write it as
∫sin(x)dx/sin2= ∫ sin(x)dx/(1- cos2(x)).
Now let u= cos(x) so du= sin(x)dx and the integeral becomes
∫ du/(1-u3). The denominator factors as (1-u)(1+ u) so- partial fractions. 1/(1-u2)= (1/2)/(1-u)+ (1/2)/(1+u).
The integral is now (1/2)∫du/(1-u)+ (1/2)∫du/(1+u)= -(1/2)ln|1-u|+ (1/2)ln|1+u|+C= (1/2)ln(|1+u|/|1-u|)+ C= (1/2)ln((1+sin(x))/(x-sin(x)))+ C= ln(√((1+sin(x))/(1-sin(x)))+ C
Hmmm, maybe Reihhard-t's solution isn't so scary after all!
actually all equation can be integrated,but for you case optimus,your further integration will be very complicated,and frankly speaking ,I don't know how to integrate more,to get exactly like my previous answer.
so I recommend you to implement my previos method.that one is actually a math trick.it is fast and incisive
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