Question

help me out.. I'm noob

Answer 1

Answer:

$\int \frac{ \cos(x) }{ \cos(x) + \sin(x) } dx$

Use substitution

$\int \frac{ \frac{1 - { \tan( \frac{x}{2} ) }^{2} }{1 + { \tan( \frac{x}{2} ) }^{2} } }{ \frac{1 - { \tan( \frac{x}{2} ) }^{2} }{1 + { \tan( \frac{x}{2} ) }^{2} } + \frac{2 \tan( \frac{x}{2} ) }{1 - { \tan( \frac{x}{2} ) }^{2} } } dx$

Substitute the differential

$\int \frac{ \frac{1 - { \tan( \frac{x}{2} ) }^{2} }{1 + { \tan( \frac{x}{2} ) }^{2} } }{ \frac{1 - { \tan( \frac{x}{2} ) }^{2} }{1 + { \tan( \frac{x}{2} ) }^{2} } + \frac{2 \tan( \frac{x}{2} ) }{1 + { \tan( \frac{x}{2} ) }^{2} } } \times \frac{1}{ \frac{1}{2}(1 + { \tan( \frac{x}{2} ) ) }^{2} } dt$

substitute 't' instead of tan(x/2)

$\int \dfrac{ \frac{1 - { t}^{2} }{1 + { t }^{2} } }{ \frac{1 - { t }^{2} }{1 + {t}^{2} } + \frac{2 t}{1 + { t}^{2} } } \times \frac{1}{ \frac{1}{2} (1 + {t}^{2} )} dt$

$\int \dfrac{ \frac{1 - { t}^{2} }{1 + { t }^{2} } }{ \frac{1 - {t}^{2} + 2 t}{1 + { t}^{2} } } \times \frac{1}{ \frac{1}{2} + \frac{ {t}^{2} }{2} } dt$

Simplify

$\int \frac{1 - {t}^{2} }{1 - {t}^{2} + 2t} \times \frac{2}{1 + {t}^{2} } dt$

Multiply the fraction

$\int \frac{2 - 2 {t}^{2} }{(1 - {t}^{2} + 2t)(1 + {t}^{2} ) }dt$

Rewrite the fraction

$\int \frac{ - t + 1}{1 - {t}^{2} + 2t} + \frac{ - t + 1}{1 + {t}^{2} } dt$

Use the properties of integrals

$\int \frac{ - t + 1}{1 - {t}^{2} + 2t} dt \int \frac{ - t + 1}{1 + {t}^{2} }dt$

Evaluate the integrals

$\frac{1}{2 } \times ln( |1 - {t}^{2} + 2t | ) - \frac{1}{2} \times ln (1 + {t}^{2} ) + \arctan(t)$

Now,

here t = tan(x/2)

$\frac{1}{2 } \times ln( |1 - { \tan( \frac{x}{2} ) }^{2} + 2 \tan( \frac{x}{2} ) | ) - \frac{1}{2} \times ln (1 + { \tan( \frac{x}{2} ) }^{2} ) + \arctan( \tan( \frac{x}{2} ) )$

$\frac{1}{2 } \times \ln( |1 - { \tan( \frac{x}{2} ) }^{2} + 2 \tan( \frac{x}{2} ) | ) - \frac{1}{2} \times \ln (1 + { \tan( \frac{x}{2} ) }^{2} ) + \frac{x}{2}$

Add the constant of integration ‘C’

$\frac{1}{2 } \times \ln( |1 - { \tan( \frac{x}{2} ) }^{2} + 2 \tan( \frac{x}{2} ) | ) - \frac{1}{2} \times \ln (1 + { \tan( \frac{x}{2} ) }^{2} ) + \frac{x}{2} + C$

Here,

$C\in \: {\mathbb{R}}$

Answer 2

Given,

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx}$

We can divide both the numerator and the denominator by $\displaystyle\sf {\cos x.}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\int\dfrac {\frac {\cos x}{\cos x}}{\frac {\cos x+\sin x}{\cos x}}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\int\dfrac {1}{1+\tan x}\ dx}$

We can write the numerator 1 as follows:

$\displaystyle\longrightarrow\sf{1=\dfrac {1+\tan x+1-\tan x}{2}}$

Then the integral becomes,

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\int\dfrac {1+\tan x+1-\tan x}{2(1+\tan x)}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {1}{2}\int\left (\dfrac {1+\tan x}{1+\tan x}+\dfrac {1-\tan x}{1+\tan x}\right)\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {1}{2}\int\left (1+\dfrac {1-\frac {\sin x}{\cos x}}{1+\frac {\sin x}{\cos x}}\right)\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {1}{2}\int\left (1+\dfrac {\frac {\cos x-\sin x}{\cos x}}{\frac {\cos x+\sin x}{\cos x}}\right)\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {1}{2}\int\left (1+\dfrac {\cos x-\sin x}{\cos x+\sin x}\right)\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {1}{2}\int1\ dx+\dfrac {1}{2}\int\dfrac {\cos x-\sin x}{\cos x+\sin x}\ dx}$

Since $\displaystyle\sf {\dfrac {d}{dx}\left [\sin x+\cos x\right]=\cos x-\sin x,}$

$\displaystyle\longrightarrow\sf{\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac {x}{2}+\dfrac {1}{2}\ln\left|\cos x+\sin x\right|+c}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\int\dfrac {\cos x}{\cos x+\sin x}\ dx=\dfrac{x+\ln\left|\cos x+\sin x\right|}{2}+c}}}$