Question

integration of root over 1 + sec x dx

Answer 1

$\int\sqrt{1+secx}\ dx$ $=2sin^{-1}(\sqrt{2} sin \frac{x}{2} )+C$

Step-by-step explanation:

$\int\sqrt{1+secx}\ dx$

$=\int\sqrt{\frac{1+cos\ x}{cos \ x} } \ dx$             $[ sec\ x =\frac{1}{cos\ x} ]$

$=\int \sqrt{\frac{2cos^2\frac{x}{2} }{1-2sin^2\frac{x}{2} } } \ dx$              $[1+cos 2x=2cos^2x \ and \ cos2x= 1-2sin^2x]$    

$=\sqrt{2} \int\frac{cos\frac{x}{2} }{\sqrt{1+sin^2\frac{x}{2} } } \ dx$

Let

$sin \frac{x}{2} =z\\\\\frac{1}{2} cos\frac{x}{2} \ dx = dz$

$=2\sqrt{2} \int\frac{dz}{\sqrt{1-2z^2}}$

$=\frac{2\sqrt{2}}{\sqrt{2}} \int\frac{dz}{\sqrt{(\frac{1}{\sqrt{2} } )^2-z^2}}$

$=2 sin ^{-1}\frac{z}{\frac{1}{\sqrt{2} } } +C$

$=2 sin ^{-1}{\sqrt{2}\ z} +C$

$=2sin^{-1}(\sqrt{2} sin \frac{x}{2} )+C$   [$sin \frac{x}{2} =z$]