Tricky?

[BigGlenntheHeavy]

$\displaystyle Evaluate \ \int csc(x)dx, \ now \ if \ you \ look \ in \ the \ front \ of \ your \ calculus \ book,$

$\displaystyle it \ tells \ you \ that \ \int csc(x)dx \ = \ -ln|csc(x)+cot(x)| \ + \ C.$

$\displaystyle But \ how \ did \ they \ derive \ this \ formula? \ Now, \ I \ will \ show \ you \ one \ way, \ perhaps$

$\displaystyle you \ have \ another?$

$\displaystyle \int csc(x)dx \ = \ \int\frac{dx}{sin(x)}. \ Now \ let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2), \ \implies \ sin(x) \ = \ \frac{2u}{1+u^2}$

$\displaystyle and \ dx \ = \ \frac{2du}{1+u^2}.$

$\displaystyle Ergo, \ \int\frac{dx}{sin(x)} \ = \ \int\frac{2du/(1+u^2)}{2u/(1+u^2)} \ = \ \int\frac{du}{u} \ = \ ln|u|+C$

$\displaystyle = \ ln\bigg|\frac{sin(x)}{1+cos(x)}\bigg|+C \ = \ -ln\bigg|\frac{1+cos(x)}{sin(x)}\bigg|+C$

$\displaystyle = \ -ln\bigg|\frac{1}{sin(x)}+\frac{cos(x)}{sin(x)}\bigg|+C \ = \ -ln|csc(x)+cot(x)|+C, \ QED.$

 

[BigGlenntheHeavy]

$\displaystyle Now \ the \ standard \ way \ to \ integrate \ \int csc(x)dx \ is:$

$\displaystyle \int csc(x)\bigg[\frac{csc(x)+cot(x)}{csc(x)+cot(x)}\bigg]dx \ = \ \int\frac{csc^2(x)+csc(x)cot(x)}{csc(x)+cot(x)}dx.$

$\displaystyle Now, \ if \ we \ let \ u \ = \ csc(x)+cot(x), \ then, \ -du \ = \ [csc^2(x)+csc(x)cot(x)]dx.$

$\displaystyle Hence, \ we \ have \ -\int\frac{du}{u} \ = \ -ln|u|+C \ = \ -ln|csc(x)+cot(x)| +C.$

$\displaystyle However \ to \ evaluate \ \int\frac{dx}{1+sin(x)-cos(x)}, \ we \ must \ use \ the \ formula \ in \ my$

$\displaystyle above \ thread.$