Question

I am trying to integrate 2(tanx)(sec^2x-1) from 0 to pi/4

Answer 1

So far I have distributed and broken up the integral into 2

Answer 2

Interesting... let's see if we can play with trigonometry to simplify things a bit...
\[\Large \int\limits_{0}^{\frac \pi 4}2\tan(x)[\sec^2(x)-1]dx\]

Answer 3

I began by taking out the constant, then distributing the tanx in and splitting up the problem

Answer 4

That's good. Do you know the integral of tan(x) ?

Answer 5

\[-\ln \left| cosx \right|\]

Answer 6

or \[\Large \ln|\sec(x)|\]
Okay, so we have no problem :)

Answer 7

\[\large \int\limits_{0}^{\frac \pi 4}2\tan(x)[\sec^2(x)-1]dx=\int\limits_{0}^{\frac \pi 4}2\tan(x)\sec^2(x)dx -\int\limits_{0}^{\frac \pi 4}2\tan(x)dx\]

Answer 8

So, can you do it from here?

Answer 9

I got \[\tan ^{2}x \] for the first integral

Answer 10

What? NO! It should be \(\large \sec^2(x)\)

Answer 11

LOL
Just messing with you :)
It could actually be either :D

Answer 12

\[\Large \sec^2(x) = \tan^2(x) + 1\]

Answer 13

lol, cool

Answer 14

Anyway, so... moving on?

Answer 15

I think I'm just having issues evaluating then :/

Answer 16

Okay, in the end we have
\[\LARGE \left.\tan^2(x)- 2\ln|\sec(x)|\right]_0^{\frac\pi4}\]

Answer 17

Or if it makes things simpler, one tiny tweak with the properties of logs...
\[\LARGE \left.\tan^2(x)- \ln|\sec^{\color{red}2}(x)|\right]_0^{\frac\pi4}\]

Answer 18

nice, good tip

Answer 19

Okay, so evaluating this...
\[\Large \tan^2\left(\frac\pi4\right)-\ln\left|\sec^2\left(\frac\pi4\right)\right|=1-\ln2\]

Answer 20

\[\Large \tan(0) - \ln |\sec^2(0)|=0-\ln(1)=0\]

Answer 21

Et voila ^_^

Answer 22

hmmm

Answer 23

ohhhh

Answer 24

I understand now

Answer 25

I need to review those special angles

Answer 26

thank you :)

Answer 27

^_^
That's good