Question

Integrate tan 2x

Answer 1

\(\tan 2x = \dfrac{\sin 2x}{\cos2x}\)

So we can use u-sub and let \(u = \cos2x,~du=-2\sin2x~dx\)

Answer 2

do a little algebra for du, so you have \(-2du = \sin 2x~dx\)

Answer 3

\(-\dfrac{1}{2} du\), not -2, sorry!

Answer 4

So \[\int\tan 2x~dx = \int\dfrac{\sin2x}{\cos2x}dx = -\dfrac{1}{2}\int\dfrac{1}{u}du\]

Answer 5

does this help? @Garadon_Shagan

Answer 6

To be honest geerky I am still a little lost... The chain rule is something I am still studying and my book is basically worthless. First off, why does u = cos 2x and not sin 2x?

Answer 7

But thanks for all the help so far! It did clear a lot of things up!

Answer 8

then du would have to be in denominator, so integrating it wouldn't make sense

Answer 9

so we are better start with u=cos 2x

Answer 10

du being the derivative of u, correct?

Answer 11

yeah, you take derivative of u in respect of x. then you multiply both sides by dx, basically.

so you have \(\dfrac{du}{dx} = -2\sin(2x) \longrightarrow dx\cdot\dfrac{du}{dx} = du = -2\sin2x~dx\)

Answer 12

so you can make these subs to make integral easier to solve

Answer 13

makes sense so far? @Garadon_Shagan

Answer 14

I'm going to have to find someplace to get a little more chain rule instruction because I can't really grasp why it is essential that u = cos 2x rather than sin 2x and how you arrive at that last integral of S 1/u I have no idea. But your answer is correct and I do have a better sense of what to do now, Thank you!

Answer 15

ever heard of Khan Academy? perhaps this helps; video about chain rule for you to have your feet on table and watch:

https://www.khanacademy.org/math/differential-calculus/taking-derivatives/chain_rule/v/chain-rule-introduction

Answer 16

I'll check it out, thanks!