Solution to Problem 11.8-6

6) The problem is to integrate

ln(tan(x)) dx sin(2x)

Step 1: Use the double angle formula to make the argument expressions of the trig functions in the numerator and denominator the same. That is, write the denominator as trig function(s) of x rather than 2x. When you've done that, click here to continue.

You should have

ln(tan(x)) dx = sin(2x)

1 2

ln(tan(x)) dx sin(x)cos(x)

Step 2: Use the definitions of tangent and secant and trig identities to find something you can multiply numerator and denominator by to turn all the trig functions here into tangents. When you've done that, click here to continue.

You should have found when you multiply the denominator by sec²(x) you can then apply the definitions of secant and tangent. That converts sin(x)cos(x)sec²(x) to tan(x). So you should have

ln(tan(x)) dx = sin(2x)

1 2

ln(tan(x)) sec²(x) dx = sin(x)cos(x)sec²(x)

1 2

ln(tan(x))(tan²(x) + 1) dx tan(x)

Step 3: Use the tangent substitution and make the cancellation that simplifies this integrand. Then click here to continue.

Your substitution is

u = tan(x)
and
du = dx u² + 1

You should have

ln(tan(x)) dx = sin(2x)

1 2

ln(u)(u² + 1) du = u(u² + 1)

1 2

ln(u) du u

Step 4: Apply simple substitution. Can you see that 1/u is the derivative of ln(u)? Apply the appropriate substitution and integrate the resulting expression, then click here to continue.

Your substitution is

v = ln(u)
and
du dv = u

You should get

ln(tan(x)) dx = sin(2x)

1 2

1 v dv = 4

v² + C

Step 5: Substitute back. You have two levels of back-substitution. Do them, and then click here to see the final answer.

The back substitution goes like this

ln(tan(x)) dx = sin(2x)

1 4

v² + C =

1 4

ln²(u) + C =

1 4

ln²(tan(x)) + C

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