Solve the integral:

$\displaystyle \int _0 ^{\pi} \frac{x~sin(x)}{1 + cos^2(x)} dx$

(Source: http://www.math.utah.edu/)

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- Apr 15th 2014, 04:49 PMtopsquarkProblem #1 - Integration
Solve the integral:

$\displaystyle \int _0 ^{\pi} \frac{x~sin(x)}{1 + cos^2(x)} dx$

(Source: http://www.math.utah.edu/) - Apr 20th 2014, 08:14 AMtopsquarkre: Problem #1 - Integration
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