Question

If the integral $\int \frac{5\tan x}{\tan x-2}dx=x+a\ln |\sin x-2\cos x|+k$, then $a$ is equal to

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

Answer: (D)

$2$

$I=\int \left(\frac{5\left(\frac{sinx}{cosx}\right)}{\left(\frac{sinx}{cosx}\right)-2}\right)dx=5\int \left(\frac{sinx}{sinx-2cosx}\right)dx$

$\therefore sinx=A(sinx-2cosx)+B\left(\frac{d(sinx-2cosx)}{dx}\right)orsinx=A(sinx-2cosx)+B(cosx+2sinx)=(A+2B)sinx+cosx(-2A+B)uponcomparision,wegetA+2B=1orA=-2B+1and-2A+B=0or-2(-2B+1)+B=04B-2+B=05B=2orB=\left(\frac{2}{5}\right)thenA=-2\left(\frac{2}{5}\right)+1=\left(\frac{-4}{5}\right)+1=\left(\frac{1}{5}\right)\therefore I=5\int \left(\frac{\left(\frac{1}{5}\right)(sinx-2cosx)+\left(\frac{2}{5}\right)(cosx+2sinx)}{sinx-2cosx}\right)dx=5[\int (\frac{1}{5})dx+\int (\frac{2}{5}\left)\right(\frac{cosx+2sinx}{sinx-2cosx}\left)dx\right]=x+2\int \left(\frac{cosx+2sinx}{sinx-2cosx}\right)dxletsinx-2cosx=tthen(cosx+2sinx)dx=dt\therefore I=x+2\int \left(\frac{dt}{t}\right)=x+2[log|t|+c]=x+2[log|sinx-2cosx|+c]\therefore a=2$