Question #434db

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Answer 1 · 2016-09-11T15:38:54

$1/2[1/sqrt2ln|sec(x-pi/4)+tan(x-pi/4)|-1/(cosx+sinx)]+C$ .

The Integrand $=secx/(1+tan x)^2=(1/cosx)/((1+sinx/cosx)^2)$

$=(1/cosx)(cos^2x/(cosx+sinx)^2)=cosx/(cosx+sinx)^2$

Now, using Quotient Rule for Diffn., we have, $d/dx(1/(cosx+sinx))$

$={(cosx+sinx)d/dx(1)-1d/dx(cosx+sinx)}/(cosx+sinx)^2$

$=(sinx-cosx)/(cosx+sinx)^2$ .

$rArr int(sinx-cosx)/(cosx+sinx)^2dx=1/(cosx+sinx)............(star)$

$I=intcosx/(cosx+sinx)^2dx=1/2int(2cosx)/(cosx+sinx)^2dx$

$=1/2int{(cosx+sinx)+(cosx-sinx)}/(cosx+sinx)^2dx$

$=1/2[int(cosx+sinx)/(cosx+sinx)^2dx+int(cosx-sinx)/(cosx+sinx)^2dx]$

$=1/2[int1/(cosx+sinx)dx-int(sin-cosx)/(cosx+sinx)^2dx]$

$=1/2[J-1/(cosx+sinx)]...........,[by (star)]$ , where,

$J=int1/(cosx+sinx)dx$

Here, $cosx+sinx=sqrt2(1/sqrt2cosx+1/sqrt2sinx)=sqrt2cos(x-pi/4)$

$:. J =int1/(sqrt2cos(x-pi/4))dx=1/sqrt2intsec(x-pi/4)dx$

$=1/sqrt2ln|sec(x-pi/4)+tan(x-pi/4)|$ .

Altogether,

$I=1/2[1/sqrt2ln|sec(x-pi/4)+tan(x-pi/4)|-1/(cosx+sinx)]+C$ .

Enjoy Maths.!