∫ (sinx+cosx)/√sin2x dx
We can write above integral as,
= \(\int\frac{sinx+cosx}{\sqrt{1-1+sin2x}}\)
[Adding and subtracting 1 in denominator]
= \(\int\frac{sinx+cosx}{\sqrt{1-(1-sin2x)}}\)dx
= \(\int\frac{sinx+cosx}{\sqrt{1-(sin^2x+cos^2x-2\,sinx\,cosx)}}\)dx
∵ sin2x + cos2x = 1 and
sin2x = 2 sinx cosx
= \(\int\frac{sinx+cosx}{\sqrt{1-(sinx- cosx)^2}}\)dx
∵ sin2x + cos2x - 2 sinx cosx
= (sinx – cosx)2
Put sinx – cosx = t
Differentiating w.r.t x we get,
(cosx + sinx)dx = dt
Putting values we get,