Home > Calculus > Integrals of Trigonometric Functions

How do you find #int csc(x)dx #?

Answer 1

#int csc(x)dx = lnabs(csc(x) - cot(x)) + C#

As is often with integrals, there are a couple of different ways to solve this. One method involves re-writing #csc(x)# as:

#csc(x) = 1/sin(x) = sin(x)/sin^2(x) = sin(x)/(1-cos^2(x))#

Using a u-substitution where #u = cos(x)# allows us to simplify things via partial fractions:

#int 1/(1-u^2)du = -1/2int (1/(1-u) + 1/(1+u))du#

Assessing the yields of integrals:

# -1/2(lnabs(u-1) - lnabs(u+1)) + C = 1/2lnabs((u-1)/(u+1)) + C#

Substituting #cos(x)# for #u# yields:

#1/2lnabs((cos(x)-1)/(cos(x)+1)) + C#

By multiplying this by 1, we can make it simpler:

#1/2lnabs((cos(x)-1)/(cos(x)+1)((cos(x)-1)/(cos(x)-1))) + C = 1/2lnabs((cos(x) − 1)^2/(1 − cos^2(x)))+ C#

Additionally, applying trig identities and logarithm properties:

#1/2lnabs(((1-cos(x))/sin(x))^2) + C = lnabs((1-cos(x))/sin(x)) + C = lnabs( 1/sin(x)-cos(x)/sin(x)) +C = lnabs(csc(x) - cot(x)) + C#

We get to our definitive response.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Aurora Alvarado

To find the integral of csc(x)dx, you can use the formula:

∫csc(x)dx = -ln|csc(x) + cot(x)| + C

where C is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

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.