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Question
Evaluate = `int (tan x)/(sec x + tan x)` . dx
Sum
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Solution
Let I = `int (tan x)/(sec x + tan x)` . dx
I = `int (sinx/cosx)/((1/cosx + sinx/cosx))` dx
= `int ((sinx/cosx))/(((1 + sinx)/cos))` dx
= `int (sinx)/(1 + sin x)` dx
= `int [sinx/1 + sinx xx (1 - sinx)/(1 - sinx)]` dx
= `int (sin x - sin^2x)/cos^2x` dx
= `int (sinx - sin^2x)/cos^2x` dx
= `int sinx/cos. 1/cosx dx - int (sin^2x)/cos^2x dx`
= `int tan x . sec x dx - int (sec^2 x -1)` dx
= `int tanx. sec x dx - [int sec^2x dx - int 1. dx]`
= sec x - tan x + x + c
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