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
Hi, masters.
Can I ask a question about definite integration?
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.
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
please.. it's pi, not pie
$\displaystyle{\int_0^\pi}x \sin(1-x^2)~dx$
$u=1-x^2, ~~du=-2x~dx$
$-\dfrac 1 2\displaystyle{\int_1^{1-\pi^2}}\sin(u)~du$
$-\dfrac 1 2 \cos(u)|^{1-\pi^2}_1$
$-\dfrac 1 2 \left(\cos(1-\pi^2)-\cos(1)\right)$
$\dfrac 1 2 \left(\cos(1)- \cos(1-\pi^2)\right)$
Your mistake is that $\cos(1) \neq 1$