Question

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

• A. $-1$
• B. $-2$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$I=\int \frac{5tanx}{tanx-2}dx$

$I=\int \frac{5\frac{sinx}{cosx}}{\frac{sinx}{cosx}-2}dx\Rightarrow \int \frac{5sinx}{sinx-2cosx}dx$

$\left[\begin{array}{c}\because 5sinx\leftrightarrow sinx-2cosx+2cosx+4sinx\end{array}\right]$

$I=\int \left(\begin{array}{c}\frac{sinx-2cosx}{sinx-2cosx}+\frac{4sinx+2cosx}{sinx-2cosx}\end{array}\right)dx$

$I=\int \left(\begin{array}{c}\frac{sinx-2cosx}{sinx-2cosx}\end{array}\right)dx+\int \left(\begin{array}{c}\frac{4sinx+2cosx}{sinx-2cosx}\end{array}\right)dx$

$I=I_1+I_2$ ..............(i)

$I_1=\int dx\Rightarrow x+c_1$

$I_2=\int \frac{2(2sinx+cosx)}{sinx-2cosx}dx$

put $sinx-2cosx=t\to (2sinx+cosx)dx=dt$

$\Rightarrow \int \frac{2}{t}dt$

$\Rightarrow 2ln\left|\begin{array}{c}t\end{array}\right|+c_2$

from (i) we get,

$I=x+2ln\left|\begin{array}{c}sinx-2cosx\end{array}\right|+C$........(ii)

$I=x+\alpha ln\left|\begin{array}{c}sinx-2cosx\end{array}\right|+C$...........(iii)

from (ii) and (iii)

$\alpha =2$

$\therefore \text{correct option is D}$