Solve Integration Problem: tan θ sec θ → (1/2)ln(3/2)

[chwala]

Homework Statement

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

Homework Equations

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?

 

[chwala]

anyone on this....

 

[haruspex]

show that $∫ tan θ sec θ dθ$

That should be sec 2θ, right?

$= ln (√3)/2- (1/2)ln (1/2)$
ok am i correct up to this point?

a little careless with the parentheses.

 

[chwala]

Yes sir sec2¤

 

[haruspex]

Yes sir sec2¤

Ok. What about my other comment?

 

[chwala]

That should be sec 2θ, right?

a little careless with the parentheses.

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

 

[chwala]

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

is that better? i have a problem using latex

 

[chwala]

is that better? i have a problem using latex

is this correct?

Homework Statement

given $tan 2θ-tan θ≡ tan θ sec 2θ$ show that $∫ tan θ sec 2θ dθ$ = $\frac 1 2$ $ln\frac 3 2$, limits are from θ= 0 to θ=π/6

Homework Equations

The Attempt at a Solution

$∫ tan 2θ-tan θ dθ$ →
-$\frac 1 2$ ln cos 2θ + ln cos θ

 

[haruspex]

is that better? i have a problem using latex

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
$= (ln (√3))/2- (1/2)ln (1/2)$

 

[chwala]

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
$= (ln (√3))/2- (1/2)ln (1/2)$

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

 

[haruspex]

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

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

 

[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...

 

[haruspex]

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...

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