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What is the integral of #int (csc2x)dx #?

Answer 1

#int csc(2x) "d"x = -1/2 ln|csc(2x) + cot(2x)| + "constant"#

#int csc(2x) "d"x = 1/2 ln|tan(x)| + "constant"#

Both answers are equivalent. Differentiate them if you have any doubt.

First, use the #sin# double angle formula #sin2theta -= 2sinthetacostheta# and the Pythagorean Identity #sin^2theta + cos^2theta -= 1#.

#int csc(2x) "d"x = int 1/sin(2x) "d"x#

#= int 1/(2sin(x)cos(x)) "d"x#

#= int cos(x)/(2sin(x)cos^2(x)) "d"x#

#= 1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x#

Next, substitute #u = sin(x)#.

#frac{"d"u}{"d"x} = cos(x)#

#frac{"d"u}{"d"x} * "d"x = cos(x)*"d"x#

#1/2 int cos(x)/(sin(x)(1-sin^2(x))) "d"x = 1/2 int 1/(u(1-u^2)) "d"u#

Next write the partial fractions.

#1/2 int 1/(u(1-u^2)) "d"u = 1/2 int 1/u "d"u #

#- 1/4 int 1/(1+u) "d"u - 1/4 int 1/(u-1) "d"u#

#= 1/4 ln|u^2/(1-u^2)| + "constant"#

Now revert #u# back to #sin(x)#.

#1/4 ln|u^2/(1-u^2)| = 1/4 ln|sin^2(x)/(1-sin^2(x))| #

#= 1/4 ln|sin^2(x)/cos^2(x)| #

#= 1/4 ln|tan^2(x)| #

#= 1/2 ln|tan(x)| #

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Answer 2

Emma Aldridge

The integral of csc^2(x) dx is -cot(x) + C, where C is the constant of integration.

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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.