Home > Calculus > Definite and indefinite integrals

How do you evaluate the integral of #int tan(x)ln(cosx) dx#?

Answer 1

Christopher Allers

#-ln^2(cosx)/2+C#

We can solve this integral using u-substitution, but it is fairly unapparent from the outset.

If we let #u=ln(cosx)#, differentiating this reveals that

#du=(d/dx(cosx))/cosxdx=(-sinx)/cosxdx=-tanxdx#

Multiply the interior and exterior of the integral by #-1#.

#inttanxln(cosx)dx=-int-tanxln(cosx)dx=-intudu#

Integrating this gives #-u^2/2+C#, and since #u=ln(cosx)#, this becomes #-ln^2(cosx)/2+C#.

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

Answer 2

Ezra Adams

To evaluate the integral ∫ tan(x) ln(cos(x)) dx, you can use integration by parts. Let u = ln(cos(x)) and dv = tan(x) dx. Then differentiate u to get du and integrate dv to get v. After that, apply the integration by parts formula:

∫ u dv = uv - ∫ v du

Then, substitute u, v, du, and dv into the formula and integrate accordingly. You may need to use trigonometric identities to simplify the expression during 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.