Question

dx/√(1+sinx). find the integral

Answer 1

Given, $\bold{\int{\frac{dx}{\sqrt{1+sinx}}}}$

First of all we have to resolve $\bold{\sqrt{1+sinx}}$
sinx = cos(π/2 - x)
∴ 1 + sinx = 1 + cos(π/2 - x)
But we know, 1 + cosФ = 2cos²Ф/2 , use it here,
∴ 1 + cos(π/2 - x) = 2cos²(π/2 - x)/2
= 2cos²(π/4 - x/2)

Hence, $\bold{\sqrt{1+sinx}=\sqrt{2}cos(\frac{\pi}{4}-\frac{x}{2})}$
So, integration converts in $\bold{\int{\frac{dx}{\sqrt{2}cos(\pi/4-x/2)}}}$
= $\bold{\frac{1}{\sqrt{2}}\int{sec(\pi/4-x/2).dx}}$
= $\bold{\frac{-2\times1}{\sqrt{2}}log|tan(\pi/4-x/2)+sec(\pi/4-x/2)|} +C$
= $\bold{-\sqrt{2}log|tan(\pi/4-x/2)+sec(\pi/4-x/2)|} +C$