Question

$\int \frac{1}{1+\tan x}dx=$

• A. $\frac{1}{2}x-\frac{1}{2}\log |\sin x+Cosx|+c$
• B. $\frac{1}{2}x+\frac{1}{2}\log |\sin x+Cosx|+c$
• C. $\frac{3}{2}x+\frac{1}{2}\log |\sin x+Cosx|+c$
• D. $\frac{3}{2}x-\frac{1}{2}\log |\sin x+Cosx|+c$

Answer 1

The correct option is B $\frac{1}{2}x+\frac{1}{2}\log |\sin x+Cosx|+c$

$\int \frac{cosx}{cosx+sinx}dx$
Let's write $N\left(x\right)=\lambda \left[D^1\right(x\left)\right]+\mu \left[D\right(x\left)\right]$
$cosx=\lambda [-sinx+cosx]+\mu [cosx+sinx]$
$\begin{array}{ccc}\lambda +\mu & =& 1\\ -\lambda +\mu & =& 0\end{array}\}\Rightarrow \lambda =\frac{1}{2};\mu =\frac{1}{2}$
$\int \frac{{}^1{/}_2\left[D^1\right(x\left)\right]{+}^1{/}_2\left[D\right(x\left)\right]}{D\left(x\right)}$
$=\frac{1}{2}\int \frac{D^1\left(x\right)}{D\left(x\right)}+\frac{1}{2}\int \frac{D\left(x\right)}{D\left(x\right)}dx$
$=\frac{1}{2}\log \left[D\right(x\left)\right]+\frac{1}{2}x+c$
$=\frac{1}{2}\log [sinx+cosx]+\frac{1}{2}x+c$