How do i split this up?

Question

anonymous · Answer

|dw:1351265649045:dw|

amistre64 · Answer

what are you trying to get to?

anonymous · Answer

Im trying to do the integral of this function

amistre64 · Answer

.... ewww  that looks rough then

amistre64 · Answer

in general:$$\frac{a-b}{a+b}=\frac{a}{a+b}-\frac{b}{a+b}$$
but i cant see how that would be all that useful at the moment

anonymous · Answer

$$\frac{ sinx - cosx }{ sinx + cosx   }$$

amistre64 · Answer

it might be useful in a by parts setup

anonymous · Answer

I know its not an inmediate integral and i cant solve it by parts nor can i use substitution so i would have to rewrite it but im not sure how

amistre64 · Answer

i think i see it

anonymous · Answer

u = sinx +cosx
-du = -cosx + sinx  dx

amistre64 · Answer

Algebra got it first tho ;)

anonymous · Answer

I have a question, how does the dU become negative is it because you change the signs of COS and SINE?

anonymous · Answer

sorry @amistre64  , thought you guys were off track...

anonymous · Answer

I know that when you have a constant the dU is either multiplied or divided by it for example

amistre64 · Answer

what is the derivative of the bottom, and how does it compare to the top?  this would relate to an log solution.
$$Dx[sinx+cosx]=cosx-sinx$$
$$cosx-sinx = -(sinx-cosx)$$

$$\frac{-}{-}*\frac{sinx-cosx}{sinx+cosx}=-\frac{cosx-sinx}{sinx+cosx}=-\frac{u'}{u}$$

anonymous · Answer

f0(x)
f(x) dx = ln jf(x)j + c

anonymous · Answer

oops

amistre64 · Answer

being off track is usually how i get on track ;)

anonymous · Answer

$$\int\limits \frac{F \prime }{ F(X)} dx = Ln IF(X)I + C $$

amistre64 · Answer

your notation is peculiar, but i can read it.  so yes

anonymous · Answer

Sorry for the I :)

But this is the formula that i need to use for this right? Sorry im a noob at this.

amistre64 · Answer

that is the formula, correct

amistre64 · Answer

the pipe character is usually above the enter key:  | 

shift + \

anonymous · Answer

$$Ln \left| Sinx + COSx \right| + C$$

amistre64 · Answer

dont forget the negative that had to be introduced to work it out

amistre64 · Answer

we had to "restructure" it to get it into a form that we could use; which introduced a constant "-1"

anonymous · Answer

Its the same thing as removing the constant from the integral right?

amistre64 · Answer

$$\int \frac{p}{q}=>\int-\frac{u'}{u}=>-\int \frac{u'}{u}$$
$$-\int \frac{u'}{u}=-(ln|u|+C)=>-ln|u|+C$$

amistre64 · Answer

yes, constants can be "pulled out"

anonymous · Answer

$$-Ln \left| Sinx - Cosx \right|$$ + C

amistre64 · Answer

might be a typo but;  sin + cos, not sin-cos

anonymous · Answer

thks

amistre64 · Answer

youre welcome