Mathematics

lf $$\displaystyle \int\frac{1}{1+\cot x}dx=_{A}$$ Iog $$|Sinx+$$ Cos $$x|+$$ Bx $$+c$$, then $$A=\ldots\ldots..,\ B=\ldots\ldots\ldots.$$,


ANSWER

$$-\displaystyle \frac{1}{2},\frac{1}{2}$$


SOLUTION
$$\displaystyle \int \frac{sin\ x}{sin\ x+cos\ x}dx$$
Lets write $$N(x)= \lambda (D^{1}(x))+\mu (D(x))$$
$$\displaystyle sin\ x\ \lambda(cos\ x-sin\ x)+\mu (sin\ x+cos\ x)$$
$$\displaystyle \lambda +\mu = 0$$
$$\displaystyle \lambda =-\frac{1}{2} ;\ \mu =\frac{1}{2} $$
$$\displaystyle -\lambda +\mu =1$$
$$\displaystyle N(x)=-\frac{1}{2} \ D^{1}(x)+\frac{1}{2} \ D(x)$$
$$\displaystyle \int \frac{N(x)}{D(x)}\ dx=-\frac{1}{2} \int \frac{D^{1}(x)}{D(X)}+\frac{1}{2} \int \frac{D(X)}{D(x)}\ dx$$
$$\displaystyle =-\frac{1}{2} \log\ (D(x))+\frac{1}{2} x+c$$
$$\displaystyle =-\frac{1}{2} \log\ \left | sin\ x+cos\ x \right |+\frac{1}{2}  x+c$$
$$\displaystyle \int \frac{1}{1+cot\ x}\ dx =-\frac{1}{2} log\ \left | sin\ x+cos\ x \right |+\frac{1}{2}  x+c$$
$$\displaystyle A= -\frac{1}{2} ;\ B= \frac{1}{2} $$
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