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,

= \(\int\frac{sinx-cosx}{\sqrt{1+sin2x-1}}\)

[Adding and subtracting 1 in denominator]

= \(\int\frac{sinx-cosx}{\sqrt{(1+sin2x)-1}}\)dx

= \(\int\frac{sinx+cosx}{\sqrt{(sin^2x+cos^2x+2\,sinx\,cosx)-1}}\)dx

∵ sin²x + cos²x = 1 and 

sin2x = 2 sinx cosx

= \(\int\frac{sinx-cosx}{\sqrt{(sinx+cosx)^2-1}}\)dx

∵ sin²x + cos²x + 2 sinx cosx 

= (sinx + cosx)²

Taking minus (-) common from numerator we get,

= \(-\int\frac{(-sinx+cosx)}{\sqrt{(sinx+cosx)^2-1}}\)dx

Put sinx + cosx = t 

Differentiating w.r.t x we get, 

(cosx - sinx)dx = dt 

Putting values we get,

We know that,

Here,

x = t and 

a = 1