Question

Evaluate ∫ (sinx+cosx)/√sin2x dx

\(\int\frac{sinx+cosx}{\sqrt{sin2x}}\)dx

Answer 1

∫ (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,