Question

integrate 1/1+tanx dx

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\int\dfrac{1}{1+tan(x)}\,dx}$

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

$\displaystyle\sf{=\int\dfrac{cos(x)}{cos(x)+sin(x)}\,dx}$

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

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

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

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

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

$\displaystyle\sf{=\dfrac{1}{2}\,x+\dfrac{1}{2}\,\ln|cos(x)+sin(x)|+C}$

$\displaystyle\sf{=\dfrac{x}{2}+\ln\sqrt{sin(x)+cos(x)}+C}$