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How do you integrate #int (csc2x)dx#?

Answer 1

Aurora Alvarado

#I=1/2ln|csc2x-cot2x|+c#

We know that,

#color(red)(intcsctheta d theta=ln|csctheta- cottheta|+c)..........(1)#

Here,

#I=int(csc2x)dx#

Let, #2x=u=>2dx=du=>dx=1/2du#

#I=intcscu*1/2du#

#=1/2intcscudu...tocolor(red)(Apply(1)#

#=1/2ln|cscu-cotu|+c ,where, u=2x#

#=1/2ln|csc2x-cot2x|+c#

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Answer 2

Cameron Alexander

#int csc2x*dx=1/2ln(tanx)+C#

#int csc2x*dx#

=#int dx/(sin2x)#

=#int dx/(2sinx*cosx)# =#int ([(sinx)^2+(cosx)^2]*dx)/(2sinx*cosx)#

=#1/2int (sinx*dx)/cosx+1/2int (cosx*dx)/sinx#

=#-1/2ln(cosx)+1/2ln(sinx)+C#

=#1/2ln(sinx/cosx)+C#

=#1/2ln(tanx)+C#

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Answer 3

Paisley Johnson

To integrate ( \int \csc^2(x) , dx ), you can use the following steps:

Recognize that the integral of ( \csc^2(x) ) is equal to ( -\cot(x) + C ), where ( C ) is the constant of integration.

So, ( \int \csc^2(x) , dx = -\cot(x) + C ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.