If-5tan x tan x-2dx-x-aln-sin x-2cos x-kthen a is equal to

L.H.S = ∫5tan x tan x 2dx =∫5·sin xcos xsin xcos x 2dx =∫5sin x sin x 2cos xdx =∫(sin x 2cos x)+2(cos x+2sin x) sin x 2cos xdx =∫1 dx+2∫cos x+2sin x sin x 2cos ...

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Standard XII

Mathematics

Logarithmic Differentiation

If∫5tan x tan...

Question

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

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Solution

The correct option is B $2$
L.H.S $= \int \frac{5 tan x}{tan x - 2} d x$
$= \int \frac{5 \cdot \frac{sin x}{cos x}}{\frac{sin x}{cos x} - 2} d x$
$= \int \frac{5 sin x}{sin x - 2 cos x} d x$
$= \int \frac{(sin x - 2 cos x) + 2 (cos x + 2 sin x)}{sin x - 2 cos x} d x$
$= \int 1 d x + 2 \int \frac{cos x + 2 sin x}{sin x - 2 cos x} d x$
$= x + 2 ln | sin x - 2 cos x | + k$
Comparing to given R.H.S
Hence the value of $a = 2$

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If ∫5tanxtanx−2dx=x+aln|sinx−2cosx|+k then a is equal to

The correct option is B 2L.H.S =∫5tanxtanx−2dx =∫5⋅sinxcosxsinxcosx−2dx =∫5sinxsinx−2cosxdx =∫(sinx−2cosx)+2(cosx+2sinx)sinx−2cosxdx =∫1 dx+2∫cosx+2sinxsinx−2cosxdx =x+2ln|sinx−2cosx|+k Comparing to given R.H.S Hence the value of a=2

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