Question

Find the integrals of the function: $\frac{cos\ x-sin\ x}{1+ sin\ 2x}$

Answer 1

Answer:

Integral of the given function is

$\frac{1}{-(cosx+sinx)}+c$

Step-by-step explanation:

Concept:

I have applied change of variable method to find integral of the given function.

$\frac{cosx-sinx}{1+sin2x}\\\\=\frac{cosx-sinx}{cos^2x+sin^2x+2.sinx.cosx}\\\\=\frac{cosx-sinx}{(cosx+sinx)^2}$

$I =\int{\frac{cosx-sinx}{1+sin2x}}\\\\=\int{\frac{cosx-sinx}{(cosx+sinx)^2}\:dx$

Take,

t = cosx+sinx

$\frac{dt}{dx}=-sinx+cosx \\\\dt = (cosx-sinx)\:dx$

Now,

$I=\int{\frac{1}{t^2}}\:dt\\\\I=\int{t^{-2}}\:dt\\\\I=\frac{t^{-1}}{-1}+c\\\\I=\frac{1}{-t}+c\\\\I=\frac{1}{-(cosx+sinx)}+c$