Home > Calculus > Integrals of Trigonometric Functions

How do you find #int sinx -1/csc(x)dx #?

Answer 1

Paisley Johnson

#C#, in which #C# is a constant.

Given: #intsinx-1/cscx \ dx#.

Notice how 1/cscx=sinx#, so the integral becomes:

#=intsinx-sinx \ dx#

#=int0 \ dx#

#=C#

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

Answer 2

Cameron Alexander

#C# where #C# is an arbitrary constant.

Recall that #cscx=1/sinx.# Rewrite the integral:

#int(sinx-1/cscx)dx=int(sinx-1/(1/sinx))dx#

#=int(sinx-sinx)dx#

#=int(0)dx=C# where #C# is an arbitrary constant.

We can easily see this is true as the derivative of any constant #C#, or #d/dx(C),# is #0.#

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

Answer 3

Christopher Allers

To find ∫ (sin(x) - 1/csc(x)) dx, you can rewrite it as ∫ (sin(x) - sin(x)) dx = ∫ 0 dx = 0. So, the integral evaluates to 0.

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.