Solve the integral:
$\displaystyle \int _0 ^{\pi} \frac{x~sin(x)}{1 + cos^2(x)} dx$
(Source: http://www.math.utah.edu/)
Solve the integral:
$\displaystyle \int _0 ^{\pi} \frac{x~sin(x)}{1 + cos^2(x)} dx$
(Source: http://www.math.utah.edu/)
Well, no one answered. It's a simple enough problem to do by integration by parts but I like the solution I found as it doesn't have to do with parts.
$\displaystyle \int_0^{\pi} \frac{x~sin(x)}{1 + cos^2 (x)} dx$
Let $\displaystyle x = \pi - z$ giving dx = - dz.
Then
$\displaystyle \int_0^{\pi} \frac{x~sin(x)}{1 + cos^2 (x)} dx = - \int _{\pi}^0 \frac{( \pi - z ) sin( \pi - z )}{1 + cos^2( \pi - z )} dz$
After some work with the addition of angles formulas:
$\displaystyle \int_0^{\pi} \frac{x~sin(x)}{1 + cos^2 (x)} dx = \int_0^{\pi} \frac{ \pi sin( z )}{1 + cos^2 (z) } dz - \int_0^{ \pi } \frac{z~sin(z)}{1 + cos^2(z)} dz$
x and z are "dummy" variables so change the z on the RHS to x and move the last term on the right to the LHS:
$\displaystyle 2 \int_0^{\pi} \frac{x~sin(x)}{1 + cos^2 (x)} dx = \int_0^{\pi} \frac{ \pi sin( z )}{1 + cos^2 (z) } dz$
The integral on the RHS is elementary and gives:
$\displaystyle \int_0^{\pi} \frac{x~sin(x)}{1 + cos^2 (x)} dx = \frac{ \pi }{2} \cdot \left ( \frac{ \pi }{2} \right ) = \left ( \frac{ \pi }{2} \right ) ^2$
-Dan