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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
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B
2
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C
1
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D
2
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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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