Question

what is this definite integral equal to?

Answer 1

\[\int\limits_{0}^{\pi/3} \sec \theta d \theta\]

Answer 2

To do this one you need a special trick:

\[\sec(\theta) =\frac{ \sec(\theta)(\sec(\theta)+\tan(\theta)) }{ \sec(\theta) + \tan(\theta) }\]

Then try differentiate the denominator and see what you get ;)

Answer 3

ahh okay so unless i knew that part then I couldn't do it? xD

Answer 4

yeah it is one of those interesting ones.
can you see what you need to do now?

Answer 5

No :S

Answer 6

use integration by part

Answer 7

oh

Answer 8

ok differentiate
\[\sec \theta + \tan \theta\]
then look at the numerator of my expression prior...

Answer 9

ohh nvm lol i was thinking integrate that for some reason xD Okay thank you!!!

Answer 10

so use substitution, correct?

Answer 11

there's no substitution. It just turns out to be a nice integral leading to a logarithm according to
\[\int\limits \frac{ f'(x) }{ f(x) } dx = \ln|f(x)| + C\]

Answer 12

if you know the special trick, that is. IOW, it's horrible :)

Answer 13

lol okay thank you again!!

Answer 14

Full proof below:
\[\sec \theta = \frac{ \sec \theta ~(\sec \theta + \tan \theta )}{ (\sec \theta + \tan \theta)} = \frac{ \sec^2 \theta + \sec \theta \tan \theta }{ \sec \theta + \tan \theta} =\frac{ \frac{ d }{ dx } (\sec \theta + \tan \theta) }{ \sec \theta + \tan \theta } \]

\[\int\limits \sec \theta ~d \theta = \int\limits \frac{ \frac{ d }{ d \theta }(\sec \theta + \tan \theta) }{ \sec \theta + \tan \theta } d \theta = \ln|\sec \theta + \tan \theta| + C\]

Answer 15

wow :O thanks!!

Answer 16

it's not too dissimilar to how you figure out the integral of tan(theta) for instance when you write it as the ratio of sine to cosine...

Answer 17

Ah okay gotcha