Question

$\int\ {\frac{1}{sinx+cos} } \, dx$

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\int\dfrac{1}{sin(x)+cos(x)}\,dx}$

$\displaystyle\rm{=\int\dfrac{1}{\sqrt{2}\left\{\dfrac{1}{\sqrt{2}}\cdot\,sin(x)+\dfrac{1}{\sqrt{2}}\cdot\,cos(x)\right\}}\,dx}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\int\dfrac{1}{cos(x)\cdot\,cos\left(\dfrac{\pi}{4}\right)+sin\left(\dfrac{\pi}{4}\right)\cdot\,sin(x)}\,dx}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\int\dfrac{1}{cos\left(x-\dfrac{\pi}{4}\right)}\,dx}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\int\,sec\left(x-\dfrac{\pi}{4}\right)\,dx}$

We know,

$\boxed{\displaystyle\green{\sf{\int\,sec(x)\,dx=\ln\left|tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)\right|+C}}}$

So,

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left\{\dfrac{1}{2}\left(x-\dfrac{\pi}{4}\right)+\dfrac{\pi}{4}\right\}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left(\dfrac{x}{2}-\dfrac{\pi}{8}+\dfrac{\pi}{4}\right)\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left(\dfrac{x}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left(\dfrac{x}{2}+\dfrac{2\pi-\pi}{8}\right)\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\right|+C}$