- #1

chwala

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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

## 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?

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