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What is #int tan(2x-3)-cot(3x) #?

Answer 1

-1/2ln(cos(2x-3)) - 1/3ln(sin(3x)) + C

#I= int(tan(2x-3) - cot(3x))dx# #I =inttan(2x-3)dx - int(cot(3x)dx# #I =int sin(2x-3)/cos(2x-3)dx - int cos(3x)/sin(3x)dx# #I=I_1 - I_2 # Let us name the two integrals as #I_1# and #I_2# and solve them separately.

#I_1 = int sin(2x-3)/cos(2x-3) dx#

Let's tackle this first.

Let #u=cos(2x-3)# Diffrentiating with respect to #x# using chain rule.

#(du)/dx = -sin(2x-3)d/dx(2x-3)# #(du)/dx = -sin(2x-3)(2)# #(du)/dx = -2sin(2x-3)# #du = -2sin(2x-3)dx# #(du)/-2 = sin(2x-3)dx# #I_1 = int 1/u ((du)/-2)# #I_1 = -1/2int (du)/u# #I_1 = -1/2ln(u)# #I_1 = -1/2 ln(cos(2x-3))#

#I_2 = intcos(3x)/sin(3x) dx# Let #u=sin(3x)# Differentiating with respect to #x# using chain rule.

#(du)/dx = cos(3x)d/dx(3x)# #(du)/dx = cos(3x)3# #(du)/dx =3cos(3x)# #(du)/3 = cos(3x)dx#

#I_2 = int1/u ((du)/3)# #I_2 = 1/3 int (du)/u# #I_2 = 1/3 ln(u)#

#I_2 = 1/3 ln(sin(3x))#

We had got till the step #I=I_1-I_2#

# -1/2ln(cos(2x-3)) - 1/3ln(sin(3x)) + C# Answer

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Answer 2

Axel Alvarez

The expression is equivalent to ( \tan(2x - 3) - \cot(3x) ).

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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.