Home > Calculus > Integrals of Trigonometric Functions

How do you find the antiderivative of #ln(cosx)tanxdx#?

Answer 1

Ezra Adams

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

We wish to find:

#intln(cosx)tanxdx#

Use substitution:

If #u=ln(cosx)#, then #du=(-sinx)/cosxdx=-tanxdx#.

Thus,

#intln(cosx)tanxdx=-intln(cosx)(-tanx)dx=-intudu#

This becomes

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

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

Answer 2

Emma Aldridge

To find the antiderivative of (\ln(\cos(x))\tan(x) ,dx), follow these steps:

Use integration by parts, with (u = \ln(\cos(x))) and (dv = \tan(x) ,dx).
Find the derivative of (u), denoted as (du), and the antiderivative of (dv), denoted as (v).
Integrate (dv) to find (v), which is (\int \tan(x) ,dx).
Differentiate (u) to find (du), which is (\frac{-\sin(x)}{\cos(x)} ,dx).
Apply the integration by parts formula: [\int u ,dv = uv - \int v ,du]
Substitute the values of (u), (v), (du), and (dv) into the integration by parts formula.
Evaluate the resulting integral.
Simplify the expression if necessary.

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.