Question

$\int \frac{senxdx}{\sqrt{sin(2x+A)+sinA}}=\int \frac{secxdx}{\sqrt{2sinxcosx+(2co{s}^{2}x-1)sinA+sinA}}=\int \frac{secxdx}{\sqrt{2cosxsin(x+A)}}$\

Answer 1

$\int \frac{secxdx}{\sqrt{sin(2x+A)+sinA}}(sinC+Sin=2sin\frac{(+)}{2}cos\frac{(-)}{2})$
$=\frac{secxdx}{\sqrt{2sin(x+A)cosx}}A=a$ For simplicity
$=\frac{1dx}{\sqrt{2}cosx\sqrt{cosx}\sqrt{sin(x+a)}}=\frac{1dx}{\sqrt{2}cosx\sqrt{cosx}\sqrt{sinxcosa+cosxsina}}$
$=\frac{dx}{\sqrt{2}cosx\sqrt{cosx}\sqrt{cosx\left(\frac{sinxcosa}{cosx}\right)+\frac{cosxsina}{cosx}}}=\frac{dx}{\sqrt{2}cosx\sqrt{cosx}\sqrt{cosx}\sqrt{tanxcosa+sina}}=\frac{dx}{\sqrt{2}co{s}^{2}x\sqrt{tanxcosa+sina}}=\frac{dx}{\sqrt{2}co{s}^{2}x\sqrt{tanxcosa+sina}}=\frac{se{c}^{2}xdx}{\sqrt{2}\sqrt{tanxcosa+sina}}$
Let tan x cos a +sin a =t
$\therefore \frac{dt}{dx}=se{c}^2$ cos a (a is constant)
$\Rightarrow se{c}^{2}xdx=\frac{dt}{cosa}=dt$ sec a.
$\therefore \frac{se{c}^{2}xdx}{\sqrt{2}\sqrt{tanxcoss+sina}}=\frac{secadt}{\sqrt{2}\sqrt{t}}=\frac{seca}{\sqrt{2}}\int t^{-1/2}dt=\frac{seca}{\sqrt{2}}\frac{t^{-1/2}+1}{(-\frac{1}{2}+1)}+C=\frac{seca}{\sqrt{2}}2\sqrt{t}+C=\sqrt{2}seca\sqrt{t}+C=\sqrt{2}seca\sqrt{tanxcosA+sinA}+C$