Given $\displaystyle cos(x)= \frac{1-u^2}{1+u^2}$ then $\displaystyle u^2$= sin(x)= u dx=
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Originally Posted by Selim Given $\displaystyle cos(x)= \frac{1-u^2}{1+u^2}$ then $\displaystyle u^2=\frac{1-cos(x)}{1+cos(x)}$ $\displaystyle sin(x)=\frac{2u}{1+u^2}$ $\displaystyle dx=\frac{2\,du}{1+u^2}$ $\displaystyle u=\frac{sin(x)}{1+cos(x)}=tan({\frac{x}{2}})$ These are the substitutions that were used for $\displaystyle \int\frac{1}{3+2cos(x)}\,dx$
Last edited by ione; Mar 21st 2010 at 10:59 PM.
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