ok what am i doing rong =/
integral of xsinx
i am using integration by parts and what i get is cosx-xcosx
but the answer is sinx-xcosx
$\displaystyle \int_{-1}^1 \sqrt{1-t^2}dt \cdot \frac{\sqrt{1-t^2}}{\sqrt{1-t^2}} = \int_{-1}^1 \frac{1}{\sqrt{1-t^2}}dt - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$
$\displaystyle = sin^{-1} (1) - sin^{-1} (-1) - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$
for
$\displaystyle \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$
set $\displaystyle sin \theta = \sqrt{1-t^2} \implies t = cos \theta$
$\displaystyle cos \theta d\theta = \frac{-t}{\sqrt{1-t^2}}d\theta$
so,
$\displaystyle \int \frac{t^2}{\sqrt{1-t^2}}dt = -\int cos^2\theta d\theta = - \frac{1}{2}\int (1+cos2\theta) d\theta$
$\displaystyle = -\frac{1}{2}\theta - \frac{1}{4} sin 2\theta$
$\displaystyle = -\frac{1}{2} cos^{-1} t - \frac{1}{4} sin 2(cos^{-1} t)$
so original is $\displaystyle = sin^{-1} (1) - sin^{-1} (-1) + \frac{1}{2}cos^{-1} t + \frac{1}{4} sin 2(cos^{-1} t) = \frac{\pi}{2}$