BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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\(\displaystyle Evaluate \ \int csc(x)dx, \ now \ if \ you \ look \ in \ the \ front \ of \ your \ calculus \ book,\)
\(\displaystyle it \ tells \ you \ that \ \int csc(x)dx \ = \ -ln|csc(x)+cot(x)| \ + \ C.\)
\(\displaystyle But \ how \ did \ they \ derive \ this \ formula? \ Now, \ I \ will \ show \ you \ one \ way, \ perhaps\)
\(\displaystyle you \ have \ another?\)
\(\displaystyle \int csc(x)dx \ = \ \int\frac{dx}{sin(x)}. \ Now \ let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2), \ \implies \ sin(x) \ = \ \frac{2u}{1+u^2}\)
\(\displaystyle and \ dx \ = \ \frac{2du}{1+u^2}.\)
\(\displaystyle Ergo, \ \int\frac{dx}{sin(x)} \ = \ \int\frac{2du/(1+u^2)}{2u/(1+u^2)} \ = \ \int\frac{du}{u} \ = \ ln|u|+C\)
\(\displaystyle = \ ln\bigg|\frac{sin(x)}{1+cos(x)}\bigg|+C \ = \ -ln\bigg|\frac{1+cos(x)}{sin(x)}\bigg|+C\)
\(\displaystyle = \ -ln\bigg|\frac{1}{sin(x)}+\frac{cos(x)}{sin(x)}\bigg|+C \ = \ -ln|csc(x)+cot(x)|+C, \ QED.\)
\(\displaystyle it \ tells \ you \ that \ \int csc(x)dx \ = \ -ln|csc(x)+cot(x)| \ + \ C.\)
\(\displaystyle But \ how \ did \ they \ derive \ this \ formula? \ Now, \ I \ will \ show \ you \ one \ way, \ perhaps\)
\(\displaystyle you \ have \ another?\)
\(\displaystyle \int csc(x)dx \ = \ \int\frac{dx}{sin(x)}. \ Now \ let \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2), \ \implies \ sin(x) \ = \ \frac{2u}{1+u^2}\)
\(\displaystyle and \ dx \ = \ \frac{2du}{1+u^2}.\)
\(\displaystyle Ergo, \ \int\frac{dx}{sin(x)} \ = \ \int\frac{2du/(1+u^2)}{2u/(1+u^2)} \ = \ \int\frac{du}{u} \ = \ ln|u|+C\)
\(\displaystyle = \ ln\bigg|\frac{sin(x)}{1+cos(x)}\bigg|+C \ = \ -ln\bigg|\frac{1+cos(x)}{sin(x)}\bigg|+C\)
\(\displaystyle = \ -ln\bigg|\frac{1}{sin(x)}+\frac{cos(x)}{sin(x)}\bigg|+C \ = \ -ln|csc(x)+cot(x)|+C, \ QED.\)