Integrate at limit 22/7-0 x tan …
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Naveen Sharma 8 years, 2 months ago
{tex}\int_\pi^0 {tanx \over sec x+ tan x}dx\\ = \int_\pi^0 {tanx \over sec x+ tan x}\times {sec x- tan x \over sec x- tan x} dx\\ = \int_\pi^0 {tanx ( sec x - tan x) \over sec ^2 x- tan^2 x} dx \\ = \int_\pi^0 {tanx ( sec x - tan x) } dx\\ = \int_\pi^0 {tanx.sec x - tan^2 x } dx\\ {/tex}
{tex}= \int_\pi^0 {tanx.sec x - ( sec^2 x-1) } dx\\ = \int_\pi^0 {tanx.sec x - sec^2 x+1 } dx\\ = \left | sec x - tanx + x \right |_\pi ^ 0 \\ = sec 0 - tan 0 + 1 - sec \pi + tan \pi - \pi \\ = 1 - 0 + 0 +1 + 0 - \pi \\ = 2 - \pi {/tex}
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