Calculus help on integrals?

Question

Can some one plz help me figure out what i did wrong?

integral [ tan 2x dx ]
= integral [ (sin 2x)/(cos 2x) ] dx

*Now I rewrote the equation so that I have u'/u

= (1/2)* integral [ (2 sin 2x)/(cos 2x) ]

*Now that I have u'/u i can use the formula ... integral (u'/u)=ln u

= (1/2) ln (cos 2x) + C


My problem is that the answer should be - (1/2) ln (cos 2x) + C . In other words negative, but I have no clue how they did this.

Answer 1

I think you are missing the neg sign
because d(cos2x) = - 2 sin2x

Answer 2

Integral [sin(2x)/cos(2x)] dx Call cos (2x) = z so -2sin(2x)dx = dz (remember that the derivatve of cos is - sin

so = -1/2Integral dz/z = -1/2 ln I z I = -1/2 ln I cos(2x) I +c

you should NEVER FORGET ABSOLUTE VALUE

Answer 3

Integration [Sin 2x / Cos 2x]dx

u = Cos 2x
du = - Sin 2x . d/dx(2x) dx = - Sin(2x) . 2 dx = - 2Sin(2x) dx

So Sin(2x)dx = - 1/2 du

Integration [Sin 2x / Cos 2x]dx = - 1/2 [1/u du] = - 1/2 ln(u) + c
= c - 1/2 ln(Cos 2x)

Answer 4

Integration is opposite technique to differentiation. while you're speaking approximately indefinite imperative it has lot many formulae to bear in ideas. For sure imperative its only arithmetical calculations. the priority point relies upon on what you would be studying and wherein area of the international. once you're making the question greater informative, it will be greater valuable. yet asking your seniors approximately it will be a physically powerful thought.