# Integration problem

1. Feb 3, 2017

### chwala

1. The problem statement, all variables and given/known data
given $tan 2θ-tan θ≡ tan θ sec 2θ$ show that $∫ tan θ sec θ dθ = (1/2 )ln (3/2)$ limits are from θ= 0 to θ=π/6

2. Relevant equations

3. The attempt at a solution
$∫ tan 2θ-tan θ dθ$
-(1/2 )ln cos 2θ + ln cos θ
→ $-1/2 ln 1/2 + ln √3/2$
$= ln (√3)/2- (1/2)ln (1/2)$
ok am i correct up to this point?

Last edited: Feb 3, 2017
2. Feb 3, 2017

### chwala

anyone on this.................

3. Feb 3, 2017

### haruspex

That should be sec 2θ, right?
a little careless with the parentheses.

4. Feb 3, 2017

### chwala

Yes sir sec2¤

5. Feb 3, 2017

### haruspex

Ok. What about my other comment?

6. Feb 3, 2017

### chwala

$ln \frac{3^1/2} 2$ -$\frac1 2$ $ln\frac 1 2$

Last edited: Feb 3, 2017
7. Feb 3, 2017

### chwala

is that better? i have a problem using latex

8. Feb 3, 2017

### chwala

is this correct?

Last edited: Feb 3, 2017
9. Feb 3, 2017

### haruspex

Yes, but all you needed to do was to correct from this:
$= ln (√3)/2- (1/2)ln (1/2)$
to
$= ln ((√3)/2)- (1/2)ln (1/2)$
so as to distinguish it from
$= (ln (√3))/2- (1/2)ln (1/2)$

10. Feb 4, 2017

### chwala

now how is this equal to $\frac 1 2$ ln$\frac 3 2$?

11. Feb 4, 2017

### haruspex

Just a bit of manipulation. How else could you write ln(√3)?

12. Feb 4, 2017

### chwala

i am getting $\frac 1 2$ ln 3-ln 2 - ($\frac 1 2$ ln 1 - $\frac 1 2$ln 2)
which gives me $\frac1 2$ ln 3- ln 2 - $\frac 1 2$ln 1 +$\frac 1 2$ ln2
this becomes $\frac 1 2$ln 3 + $\frac 1 2$ln 2 - ln 2
this becomes
$\frac 12$ln 3 +ln$\frac {2^1/2} 2$
$\frac 1 2$ln 3 + ln ${2^-1/2}$
$\frac 1 2$ln 3 - $\frac 1 2$ ln 2
$\frac 1 2$ln $\frac 3 2$

is this now correct? greetings from Africa............

Last edited: Feb 4, 2017
13. Feb 4, 2017

### haruspex

Yes, but you can get there a little faster.
$\frac 1 2$ln 2 - ln 2=$-\frac 1 2$ln 2