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Indefinite Integrals

Section: Mathematics

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Question : 85 of 90

Marks: +1, -0

If the

\int\; \frac{5 \tan x}{\tan x-2} dx = x + a \ln | \sin \, x - 2 \cos x | + k

a

$-1$
$-2$
$1$
$2$

Solution:

\int\; \frac{5 \tan x}{\tan x-2} dx

= \int\; \frac{5 \frac{\sin \, x}{\cos x}}{\frac{\cos x}{\cos x} - 2} dx

= \int\left( \; \frac{5 \sin \, x}{\cos x} \times \; \frac{\cos x}{\sin \, x - 2 \cos x} \right) dx

= \int\; \frac{5 \sin \, x \, dx}{\sin \, x - 2 \cos x}

= \int\left( \; \frac{4 \sin \, x + \sin \, x + 2 \cos x - 2 \cos x}{\sin \, x - 2 \cos x} \right) dx

= \int\; \frac{(\sin \, x - 2 \cos x) + (4 \sin \, x + 2 \cos x)}{\sin \, x - 2 \cos x} dx

= \int\; \frac{(\sin \, x - 2 \cos x) + 2(\cos x + 2 \sin \, x)}{(\sin \, x - 2 \cos x)} dx

= \int\; \frac{\sin \, x - 2 \cos x}{\sin \, x - 2 \cos x} dx + 2 \int\left( \; \frac{\cos x + 2 \sin \, x}{\sin \, x - 2 \cos x} \right) dx

= \int dx + 2 \int\; \frac{\cos x + 2 \sin \, x}{\sin \, x - 2 \cos x} dx

= I_1 + I_2

where

I_1 = \int dx

and

I_2 = 2 \int\; \frac{\cos x + 2 \sin \, x}{\sin \, x - 2 \cos x} dx

Put

\sin \, x - 2 \cos x = t

\; \Rightarrow (\cos x + 2 \sin \, x) dx = dt

\; \therefore I_2 = 2 \int\; \frac{dt}{t} = 2 \ln t + C

\; = 2 \ln (\sin \, x - 2 \cos x) + C

Hence,

I_1 + I_2

\; = \int dx + 2 \ln (\sin \, x - 2 \cos x) + c

\; = x + 2 \ln |(\sin \, x - 2 \cos x)| + k

\; \Rightarrow a = 2

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