# Maclaurin series for integrals

## Homework Statement

Heres the question: write the first three nonzero terms (Maclaurin Series)

$$\int^{1+sinx}_{1}(1/(sqrt{(1+ln(x))})dx$$

## The Attempt at a Solution

so for the similar questions, i use maclaurin series for common functions (you can see it from this website http://mathworld.wolfram.com/MaclaurinSeries.html" [Broken])
for this question, i tried to start with lnx, but from what i searched online, there is no maclaurin series for lnx (because it is not defined at 0), so now i am stuck. also, i tried to inegrate just the function itself, but it seems impossible.
if lnx does not have a maclaurin series, that function does not have a maclaurin series?

Last edited by a moderator:

lanedance
Homework Helper
Hi bobthetomato - welcome to PF ;)

I think the notation is a little confusing, does it help if you re-write in terms of a dummy variable?
$$f(x) = \int^{1+sinx}_{1} \frac{du}{\sqrt{1+ln(u)}}$$

note the intergal is over u from 1 to 1 + sin(x), so the ln(u) doesn't blow up

then you can differntiate w.r.t. x to find each term, which could be a little tricky...

thank your lanedance, that function looks better now.
i dont understand, why do i need to differentiate it? can i integrate the whole thing? but if i integrate the equation, i wont get maclaurin series (first three terms).

this is what i tried:

since there is a series for ln(1-x) = -x-(x^2/2)-...

let x=1-t

ln(1-(1-t))=-(1-t)-((1-t)^2/2)-...

ln(t)=-(1-t)-((1-t)^2/2)-...

but if i continue to do it, it becomes complicated and hard to differentiate

1+ln(t)= t-((1-t)^2/2)-...

1/(1+ln(t))^(1/2)= 1/(t-((1-t)^2/2)-...)^(1/2)

i stopped here because i do not know how to integrate this function.

Can someone please help me? This is only a university first year math question, so I think it shouldn't be that hard.

Here is the question again...
$$f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}$$

so i think substitution is the right place to start.

as the range of integration is near 1 (because 1+sinx can range from 1 to 2), ln(t) will not blow up.

so i set x=(1-t)

$$\ln(1-x) = -x - \frac{x^{2}}{2} - \frac{x^{3}}{3}+...$$

if i substitute 1-t into all the x

$$\ln(1-(1-t)) = -(1-t) - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...$$

and then it turns out like this

$$\ln(t) = -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...$$

now if I add 1 to the whole equation

$$\ 1 + ln(t) = 1 -1+t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...$$

which will equal to

$$\ 1 + ln(t) = t - \frac{(1-t)^{2}}{2} - \frac{(1-t)^{3}}{3}+...$$

and now if i square root the whole thing and 1/the whole equation, i will get the this...

$$f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{t-\frac{(1-t)^{2}}{2}-\frac{(1-t)^{3}}{3}+...}}$$

so know i have made it to Maclaurin Series like, i do not know what to do next. How do you simplify and integrate this?

lanedance
Homework Helper
sorry if is miss something, tex doesn't display well on this computer..
so you need to evaluate the macluarin sereis of:
$$f(x) = \int^{1+sinx}_{1} \frac{dt}{\sqrt{1+ln(t)}}$$

so rather than substituting, expanding etc. why not just evaulate the derivatives
f'(x), f''(x) and evaluate at x = 0. f(0) should be obvious

As an example, consider:
$$F(x) = \int_0^x du u^2 = \frac{u^3}{3}|_0^x = \frac{x^3}{3}$$
$$F'(x) = f(x) = \frac{d}{dx}\int_0^x du u^2 = \frac{d}{dx}\frac{x^3}{3}=x^2$$
$$F''(x) = f'(x) = ...$$
now all these should be easy to evaluate at x=0, how can you write the maclaurin expansion using those terms? (though its a very trivial expnasion for a single polynomial, but its the method that is important here)

lanedance
Homework Helper
PS if its still not clear have a look at the definition of maclaurin series in terms of derivatives of the function