Question

lf $\int \frac{1}{1+\cot x}dx=A$ $\log |\sin x+$ cos $x|+$ Bx $+c$, then $A=\dots \dots ..,B=\dots \dots \dots .$,

• A. $-\frac{1}{2},\frac{1}{2}$
• B. $-1,1$
• C. $\frac{1}{3},\frac{1}{2}$
• D. $\frac{-1}{3},\frac{1}{2}$

Answer 1

The correct option is A $-\frac{1}{2},\frac{1}{2}$

$\int \frac{1}{1+cotx}dx$
$\int \frac{sinx}{sinx+cosx}dx$
Lets write, $N\left(x\right)=\lambda \left(D^1\right(x\left)\right)+\mu \left(D\right(x\left)\right)$
$sinx=\lambda (cosx+sinx)+\mu (sinx+cosx)$
$\begin{array}{ccc}\lambda +& \mu & =0\\ -\lambda +& \mu & =1\end{array}\lambda =\frac{1}{2}\mu =\frac{-1}{2}$
$N\left(x\right)=\frac{1}{2}\left[D^1\right(x\left)\right]+{}^1{/}_2\left[D\right(x\left)\right]$
$\int \frac{N\left(x\right)}{D\left(x\right)}dx=\frac{-1}{2}\int \frac{D^1\left(x\right)}{D\left(x\right)}+\frac{1}{2}\int \frac{D\left(x\right)}{D\left(x\right)}dx$
$=-\frac{1}{2}\log \left(D\right(x\left)\right)+\frac{1}{2}x+c$
$=-\frac{1}{2}\log |sinx+cosx|+\frac{1}{2}x+c$
$A=-\frac{1}{2};B=\frac{1}{2}$