Question

# If the integral $$\displaystyle \int { \dfrac { 5\tan { x } }{ \tan { x } -2 } } dx=x+a\log { \left| \sin { x } -2\cos { x } \right| +k }$$, then $$a$$ is equal to

A
1
B
2
C
1
D
2

Solution

## The correct option is D $$2$$$$\frac { 5tanx }{tanx-2}=\frac { 5sinx }{ sinx-2cosx }$$$$\frac { d(sinx-2cosx) }{ dx }=cosx+2sinx$$$$5sinx=(sinx-2cosx)+2(cosx+2sinx)$$$$\int { \frac { 5sinx }{ sinx-2cosx }dx } =\int {\frac {(sinx-2cosx)+2(cosx+2sinx)} {sinx-2cosx}dx}$$$$= x+\int { \frac {(2(cosx+2sinx)} {sinx-2cosx}dx }$$$$taking\space sinx-2cosx=t$$$$=x+\int { 2(1/t)dt } =x+2\ln { t }+k=x+2 \ln { \left| {sinx-2cosx} \right| }+k$$$$\therefore a=2$$Mathematics

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