# Math Help - confused

1. ## confused

If $f'(x)=cos(x^2-1) and f(-1)=1.5, then f(5)= ?$

so far i have:

$\int_{-1}^5 cos(x^2-a) dx= f(5)-f(-1)$

i let $u=x^2-1$
$du=2x$

change the limits from x=5 to u=24, x=-1 to u=0

and have
$1/2 \int_0^{24} cos(u) du + f(-1)$

which gives me -1.788 but that isn't any of the answers provided!

2. Originally Posted by frog09
If $f'(x)= cos(x^2-1) and f(-1)=1.5, then f(5)= ?$

so far i have:

$\int_{-1}^5 cos(x^2-a) dx= f(5)-f(-1)$

i let $u=x^2-1$
$du=2x$ Mr F says: du = 2x dx =>dx = du/(2x). This will lead nowhere for you because the x will cause trouble. The integral cannot in fact be found using a finite number of elementary functions.

change the limits from x=5 to u=24, x=-1 to u=0 and have

$1/2 \int_0^{24} cos(u) du + f(-1)$ Mr F says: Wrong. What happened to the 1/x part of du ....?

which gives me -1.788 but that isn't any of the answers provided!
$\int_{-1}^5 \cos(x^2-1) \, dx = f(5) - f(-1)$

$\Rightarrow f(5) = \int_{-1}^5 \cos(x^2-1) \, dx + f(-1)$

$= \int_{-1}^5 \cos(x^2-1) \, dx + 1.5$

$= 1.5244 + 1.5$ correct to four decimal places using my TI-89

= 3.0244.

3. i don't know what i am doing wrong with this portion of the problem:
$\int_{-1}^5 cos(x^2-1) dx + 1.5$

i keep getting by antiderving cos
$\int_{-1}^5 sin(x^2-1)$

and when i evaluate that between -1 and 5
$sin(5^2-1)-sin({-1}^2-1)=sin(24)-sin(0)$
i get -.9056 which when added to 1.5 isn't correct. could you guide me as to where i went awry? thanks. =/

4. Originally Posted by frog09
i don't know what i am doing wrong with this portion of the problem:
$\int_{-1}^5 cos(x^2-1) dx + 1.5$

i keep getting by antiderving cos
$\int_{-1}^5 sin(x^2-1)$

and when i evaluate that between -1 and 5
$sin(5^2-1)-sin({-1}^2-1)=sin(24)-sin(0)$
i get -.9056 which when added to 1.5 isn't correct. could you guide me as to where i went awry? thanks. =/
I have already told you in my earlier reply that $\int \cos(x^2-1) \, dx$ cannot be found in terms of a finite number of elementary functions. I don't know why you're insisting on trying to integrate it.

$\int \cos(x^2-1) \, dx \neq \sin (x^2 - 1)$ .... if you bother to differentiate $\sin (x^2 - 1)$ (using the chain rule) you will quickly realise this.