Note that if we use the absolute value function and create a new function
$\text{ln}\phantom{\rule{0.2em}{0ex}}\left|x\right|,$ we can extend the domain of the natural logarithm to include
$x<0.$ Then
$\left(d\text{/}\left(dx\right)\right)\text{ln}\phantom{\rule{0.2em}{0ex}}\left|x\right|=1\text{/}x.$ This gives rise to the familiar integration formula.
Integral of (1/
u )
du
The natural logarithm is the antiderivative of the function
$f\left(u\right)=1\text{/}u\text{:}$
Although we have called our function a “logarithm,” we have not actually proved that any of the properties of logarithms hold for this function. We do so here.
Use
$u\text{-substitution}$ on the last integral in this expression. Let
$u=t\text{/}a.$ Then
$du=\left(1\text{/}a\right)dt.$ Furthermore, when
$t=a,u=1,$ and when
$t=ab,u=b.$ So we get
Thus
$\text{ln}\left({x}^{r}\right)=r\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}x$ and the proof is complete. Note that we can extend this property to irrational values of
$r$ later in this section.
Part iii. follows from parts ii. and iv. and the proof is left to you.
□
Using properties of logarithms
Use properties of logarithms to simplify the following expression into a single logarithm:
Now that we have the natural logarithm defined, we can use that function to define the number
$e.$
Definition
The number
$e$ is defined to be the real number such that
$\text{ln}\phantom{\rule{0.2em}{0ex}}e=1.$
To put it another way, the area under the curve
$y=1\text{/}t$ between
$t=1$ and
$t=e$ is
$1$ (
[link] ). The proof that such a number exists and is unique is left to you. (
Hint : Use the Intermediate Value Theorem to prove existence and the fact that
$\text{ln}\phantom{\rule{0.2em}{0ex}}x$ is increasing to prove uniqueness.)
The number
$e$ can be shown to be irrational, although we won’t do so here (see the Student Project in
Taylor and Maclaurin Series ). Its approximate value is given by
$e\approx 2.71828182846.$
The exponential function
We now turn our attention to the function
${e}^{x}.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by
$\text{exp}\phantom{\rule{0.2em}{0ex}}x.$ Then,
divide by tan^2 x
giving
1=csc^2 x -1/tan^2 x,
rewrite as:
1=1/sin^2 x -cos^2 x/sin^2 x,
multiply by sin^2 x giving:
sin^2 x=1-cos^2x.
rewrite as the familiar
sin^2 x + cos^2x=1
QED
sin x - sin x cos^2 x
sin x (1-cos^2 x)
note the identity:sin^2 x + cos^2 x = 1
thus, sin^2 x = 1 - cos^2 x
now substitute this into the above:
sin x (sin^2 x), now multiply, yielding:
sin^3 x
Q.E.D.
Andrew
take sin x common. you are left with 1-cos^2x which is sin^2x. multiply back sinx and you get sin^3x.
navin
Left side=sinx-sinx cos^2x
=sinx-sinx(1+sin^2x)
=sinx-sinx+sin^3x
=sin^3x
thats proved.