# Integration problem

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## 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

## 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?

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## Answers and Replies

Gold Member
anyone on this.................

haruspex
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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.

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Yes sir sec2¤

haruspex
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Yes sir sec2¤
Ok. What about my other comment?

Gold Member
That should be sec 2θ, right?

a little careless with the parentheses.
##ln \frac{3^1/2} 2## -##\frac1 2## ##ln\frac 1 2##

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Gold Member
##ln \frac{3^1/2} 2## -##\frac1 2## ##ln\frac 1 2##
is that better? i have a problem using latex

Gold Member
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

## The Attempt at a Solution

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

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haruspex
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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)##

Gold Member
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
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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)?

Gold Member
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............

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haruspex
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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

chwala