Question

$\int\limits {\frac{1}{\sqrt{1+cosx}} } \, dx$ =

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \int \frac{1}{ \sqrt{1 + \cos(x) } } \: dx$

$\displaystyle = \int \frac{1}{ \sqrt{2 \cos^{2} \left( \dfrac{x}{2} \right) } } \: dx$

$\displaystyle = \int \frac{1}{ \sqrt{2 } \cdot\cos \left( \dfrac{x}{2} \right) } \: dx$

$\displaystyle = \dfrac{1}{ \sqrt{2} } \int \sec \left( \dfrac{x}{2} \right) \: dx$

$\bigstar \: \: \: \bf{Put \: \: \: \dfrac{x}{2} = t }$

$\bf{ \mapsto \: \: \dfrac{1}{2}dx = dt }$

$\bf{ \mapsto \: \: dx = 2 \: dt }$

So,

$\displaystyle = \dfrac{1}{ \sqrt{2} } \int \sec \left(t \right) \cdot 2\: dt$

$\displaystyle = \dfrac{2}{ \sqrt{2} } \int \sec \left(t \right) \: dt$

$\displaystyle = \sqrt{2} \int \sec \left(t \right) \: dt$

$\displaystyle = \sqrt{2} \: \: \ln| \sec (t) + \tan(t) | + c$