Question

lf $\int \frac{1}{1+\cot x}dx{=}_A$ Iog $|Sinx+$ 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

Answer: (A)

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

$\int \frac{sinx}{sinx+cosx}dx$
Lets write $N\left(x\right)=\lambda \left(D^1\right(x\left)\right)+\mu \left(D\right(x\left)\right)$
$sinx\lambda (cosx-sinx)+\mu (sinx+cosx)$
$\lambda +\mu =0$
$\lambda =-\frac{1}{2};\mu =\frac{1}{2}$
$-\lambda +\mu =1$
$N\left(x\right)=-\frac{1}{2}{D}^1\left(x\right)+\frac{1}{2}D\left(x\right)$
$\int \frac{N\left(x\right)}{D\left(x\right)}dx=-\frac{1}{2}\int \frac{D^1\left(x\right)}{D\left(X\right)}+\frac{1}{2}\int \frac{D\left(X\right)}{D\left(x\right)}dx$
$=-\frac{1}{2}\log \left(D\right(x\left)\right)+\frac{1}{2}x+c$
$=-\frac{1}{2}\log |sinx+cosx|+\frac{1}{2}x+c$

$\int \frac{1}{1+cotx}dx=-\frac{1}{2}log|sinx+cosx|+\frac{1}{2}x+c$
$A=-\frac{1}{2};B=\frac{1}{2}$