Integral

[BigGlenntheHeavy]

$\displaystyle Evaluate: \ \int\frac{dx}{1+sin(x)-cos(x)}$

$\displaystyle Let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2) \ implies \ sin(x) \ = \ \frac{2u}{1+u^2}, \ cos(x) \ = \ \frac{1-u^2}{1+u^2}, \ and \ dx \ = \ \frac{2du}{1+u^2}$

$\displaystyle Hence, \ \int\frac{dx}{1+sin(x)-cos(x)} \ = \ \int\frac{2du/(1+u^2)}{1+[2u/(1+u^2)]-[(1-u^2)/(1+u^2)]}$

$\displaystyle = \ \int\frac{2du}{(1+u^2)+2u-(1-u^2)} \ = \ \int\frac{2du}{2(u+u^2)} \ = \ \int\frac{du}{u+u^2}$

$\displaystyle = \ \int\bigg[\frac{1}{u}-\frac{1}{1+u}\bigg]du, \ Partial \ fractions, \ = \ ln|u|-ln|1+u|+C$

$\displaystyle = \ ln\bigg|\frac{u}{1+u}\bigg|+C=ln\bigg|\frac{sin(x)/[1+cos(x)]}{1+[sin(x)/(1+cos(x)]}\bigg|+C \ = \ ln\bigg|\frac{sin(x)}{1+sin(x)+cos(x)}\bigg|+C$

$\displaystyle However, \ after \ such \ grunt \ work, \ a \ check \ is \ in \ order \ (more \ grunt \ work), \ to \ wit:$

$\displaystyle D_x\bigg[ln\bigg|\frac{sin(x)}{1+sin(x)+cos(x)}\bigg|+C\bigg] \ = \ \frac{cos(x)+1}{sin(x)+sin^2(x)+sin(x)cos(x)}$

$\displaystyle = \ \frac{cos(x)+1}{sin(x)[1+cos(x)]+[1-cos^2(x)]} \ = \ \frac{cos(x)+1}{sin(x)[1+cos(x)]+[1+cos(x)][1-cos(x)]}$

$\displaystyle = \ \frac{1+cos(x)}{[1+cos(x)][sin(x)+1-cos(x)]} \ = \ \frac{1}{1+sin(x)-cos(x)}, \ QED$