1. ## Definite integration

Hi, masters.

Integration from 0 to pie (xsin(1-x^2)dx)

I substituted (1-x^2) = t, so -1/2 integration (sint dt).
I changed the range to from 1 to 1 - pie^2 as I used substitution.

So, finally I got the answer -1/2(cos(1) - cos(1-pie^2)).
But, the answer is a bit different 1/2(1 - cos(1-pie^2)).

Can I ask what I did wrong?

Thanks a lot.

2. ## Re: Definite integration

ur answer is correct -0.69056 approx value
from books answer it is coming 0.92 which is wrong
for verification check
integrate xsin(1-x^2) from 0 to pi - Wolfram|Alpha

3. ## Re: Definite integration

$\displaystyle{\int_0^\pi}x \sin(1-x^2)~dx$

$u=1-x^2, ~~du=-2x~dx$

$-\dfrac 1 2\displaystyle{\int_1^{1-\pi^2}}\sin(u)~du$

$-\dfrac 1 2 \cos(u)|^{1-\pi^2}_1$

$-\dfrac 1 2 \left(\cos(1-\pi^2)-\cos(1)\right)$

$\dfrac 1 2 \left(\cos(1)- \cos(1-\pi^2)\right)$

Your mistake is that $\cos(1) \neq 1$

4. ## Re: Definite integration

Originally Posted by romsek

$\displaystyle{\int_0^\pi}x \sin(1-x^2)~dx$

$u=1-x^2, ~~du=-2x~dx$

$-\dfrac 1 2\displaystyle{\int_1^{1-\pi^2}}\sin(u)~du$

$-\dfrac 1 2 \cos(u)|^{1-\pi^2}_1$

$-\dfrac 1 2 \left(\cos(1-\pi^2)-\cos(1)\right)$

$\dfrac 1 2 \left(\cos(1)- \cos(1-\pi^2)\right)$

Your mistake is that $\cos(1) \neq 1$
"Pie" is also a number, \displaystyle \begin{align*} \pi \, e \approx 3.14159 \cdot 2.71828 = 8.53973 \end{align*} :P

5. ## Re: Definite integration

Sorry, but -1/2 integration sin(u)du isn't 1/1 cos(u)