# another tirg integral

• Nov 4th 2007, 08:21 PM
xslim12
another tirg integral
ok what am i doing rong =/

integral of xsinx
i am using integration by parts and what i get is cosx-xcosx
• Nov 4th 2007, 08:27 PM
kalagota
Quote:

Originally Posted by xslim12
ok what am i doing rong =/

integral of xsinx
i am using integration by parts and what i get is cosx-xcosx

IBP is
$\int udv = uv - \int vdu$
Ü
• Nov 4th 2007, 08:35 PM
xslim12
u=x du=1
dv=sinx v=-cosx

and i still get the same answer i got before =/
• Nov 4th 2007, 09:56 PM
kalagota
$\int_{-1}^1 \sqrt{1-t^2}dt \cdot \frac{\sqrt{1-t^2}}{\sqrt{1-t^2}} = \int_{-1}^1 \frac{1}{\sqrt{1-t^2}}dt - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$

$= sin^{-1} (1) - sin^{-1} (-1) - \int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$

for
$\int_{-1}^1 \frac{t^2}{\sqrt{1-t^2}}dt$

set $sin \theta = \sqrt{1-t^2} \implies t = cos \theta$
$cos \theta d\theta = \frac{-t}{\sqrt{1-t^2}}d\theta$

so,
$\int \frac{t^2}{\sqrt{1-t^2}}dt = -\int cos^2\theta d\theta = - \frac{1}{2}\int (1+cos2\theta) d\theta$

$= -\frac{1}{2}\theta - \frac{1}{4} sin 2\theta$

$= -\frac{1}{2} cos^{-1} t - \frac{1}{4} sin 2(cos^{-1} t)$

so original is $= sin^{-1} (1) - sin^{-1} (-1) + \frac{1}{2}cos^{-1} t + \frac{1}{4} sin 2(cos^{-1} t) = \frac{\pi}{2}$
• Nov 5th 2007, 04:26 AM
Krizalid
Quote:

Originally Posted by kalagota
$\int_{-1}^1 \sqrt{1-t^2}dt$

Of course, over the simetric interval $[-1,1],$ we're integratin' an even function, so we can set

$\int_{-1}^1\sqrt{1-t^2}\,dt=2\int_0^1\sqrt{1-t^2}\,dt.$