InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Evaluate `intsqrt(cotx)dx`. |
|
Answer» Put `cotx=t^(2) " so that " -" coscec "^(2)x dx=2tdtordx=(-2t)/((1+t^(4)))dt`. `thereforeintsqrt(cotx)dx=-int(2t^(2))/((t^(4)+1))dx` `=-int([(t^(2)+1)+(t^(2)-1)])/((t^(4)+1))dt=-int((t^(2)+1))/((t^(4)+1))dt-int((t^(2)-1))/((t^(4)+1))dt` `=- int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt-int((1-(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt` `=-int((1+(1)/(t^(2))))/([(t-(1)/(t))^(2)+2])dt-int((1-(1)/(t^(2))))/([(t+(1)/(t))^(2)-2])dt` `=-int(du)/([u^(2)+(sqrt(2))^(2)])-int(dv)/([v^(2)-(sqrt(2))^(2)])` `["putting"(t-(1)/(t))=u " in the 1 st integral , and "(t+(1)/(t))=v "in the 2nd"]` `=-(1)/(sqrt(2))tan^(-1)((u)/(sqrt(2)))-(1)/(2sqrt(2))log|(v-sqrt(2))/(v+sqrt(2))|+C` `=-(1)/(sqrt(2))tan^(-1)((t-(1)/(t))/(sqrt(2)))-(1)/(2sqrt(2))log|(t+(1)/(t)-sqrt(2))/(t+(1)/(t)+sqrt(2))|+C` `=-(1)/(sqrt(2))tan^(-1)((t^(2)-1)/(sqrt(2)t))-(1)/(2sqrt(2))log|(t^(2)-sqrt(2)t+1)/(t^(2)+sqrt(2)t+1)|+C` `=-(1)/(sqrt(2))tan^(-1)((cotx-1)/(sqrt(2cotx)))-(1)/(2sqrt2)log|(cotx-sqrt(2 cotx)+1)/(cotx+sqrt(2cotx)+1)|+C`. |
|