Question

Find the integrals of the function: $\frac{cos\ 2x}{(cos\ x\ +sin\ x)^2}$

Answer 1

Answer:

log(cosx+sinx)+c

Step-by-step explanation:

Concept:

I have applied decomposition method to solve this problem.

In decomposition method, the given non-integrable function is decomposed into integrable function by using algebraic identities, trigonometric identities, etc.

$\frac{cos2x}{(cosx+sinx)^2}\\\\=\frac{cos^2x-sin^2x}{(cosx+sinx)^2}\\\\=\frac{(cosx-sinx)(cosx+sinx)}{(cosx+sinx)^2}\\\\=\frac{cosx-sinx}{cosx+sinx}$

$Let,\\\\I=\int{\frac{cos2x}{(cosx+sinx)^2}}\:dx\\\\=\int{\frac{cosx-sinx}{cosx+sinx}}\:dx$

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

$Now,\\\\I=\int{\frac{1}{t}}dt\\\\I=logt+c\\\\I=log(cosx+sinx)+c$