Question

integral
$\cos( {x}^{2} )$
​

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.

\begin{lgathered}\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}\end{lgathered}

(cosx+sinx)

2

cos2x

=

(cosx+sinx)

2

cos

2

x−sin

2

x

=

(cosx+sinx)

2

(cosx−sinx)(cosx+sinx)

=

cosx+sinx

cosx−sinx

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

Let,

I=∫

(cosx+sinx)

2

cos2x

dx

=∫

cosx+sinx

cosx−sinx

dx

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

Take,

t=cosx+sinx

dx

dt

=−sinx+cosx

dt=(cosx−sinx)dx

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

Now,

I=∫

t

1

dt

I=logt+c

I=log(cosx+sinx)+c