∫(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