Help with integration by substitution

[tanhauser]

The question goes: Use the substitution t = π/4 -u to show that∫ ln(1+tan(t)) dt (for values of t between π/4 and 0) = πln(2)/8

My thoughts so far: du/dx = -1, -du=dx. By replacing t with π/4 - u the equation becomes -∫ ln(1+tan(π/4 - u)) du, using the trig identity it becomes -∫ ln(2/1+tan(u)) du.

This is as far as I got since I've been unable to integrate this function. Is there some way to manipulate the variables using various identities or is there something way obvious that I've not spotted?

I'm having a brain freeze so any help would be very much appreciated.

 

[Deleted member 4993]

The question goes: Use the substitution t = π/4 -u to show that∫ ln(1+tan(t)) dt (for values of t between π/4 and 0) = πln(2)/8

My thoughts so far: du/dx = -1, -du=dx. where did 'x' come from?

By replacing t with π/4 - u the equation becomes

-∫ ln(1+tan(π/4 - u)) du, using the trig identity it becomes -∫ ln(2/1+tan(u)) du. ← How did you get that? Use proper grouping symbols

This is as far as I got since I've been unable to integrate this function. Is there some way to manipulate the variables using various identities or is there something way obvious that I've not spotted?

I'm having a brain freeze so any help would be very much appreciated.

tan (A-B) = [tan(A) - tan(B)]/[1 + tan(A)*tan(B)}

so

tan(π/4 - u) = [1- tan(u)]/[1+tan(u)]

ln[1 + tan(π/4 - u)] = ln{2/[1+tan(u)]} = ln(2) - ln[1+tan(u)]

Now what.....(write out your original integrand with limits and this transformed integrand with limits and equate)