Mathematics

# 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

$2$

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

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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