Saved Bookmarks
| 1. |
The area of the region bounded by the curves `y=sqrt[[1+sinx]/cosx]` and `y=sqrt[[1-sinx]/cosx]` bounded by the lines x=0 and `x=pi/4` is |
|
Answer» `sinx=(2tanx/2)/(1+tan^2x/2)` `cosx=(1-tan^2x/2)/(1+tan^2x/2)` `y=sqrt((1+((2tanx/2)/(1+tan^2x/2))/((1-tan^2x/2)/(1+tan^2x/2)))` `y=(1+tanx/2)/(sqrt(1-tan^2x/2)` `y=(1-tanx/2)/sqrt(1-tan^2x)` `I=int_0^(pi/4)sqrt((1+sinx)/(cosx))-sqrt((1-sinx)/(cosx))dx` `I=int_0^(pi/4)(2tan(x/2))/sqrt(1-tan^2x/2)dx` `I=int_0^(sqrt2-1)(2t2dt)/(sqrt(1-t^2)(1+t^2))` `I=int_0^(sqrt2-1)(4dt)/((1+t^2)sqrt(1-t^2))` option b is corect. |
|