If I = ∫ (sin x)/(3sin x + cos x + 2)dx and J = ∫ (cos x)/(3sin x +...

I = ∫ \dfrac{sin x}{3sin x + cos x + 2}dx

and

J = ∫ \dfrac{cos x}{3sin x + cos x + 2}dx

, then

3J - I

is equal to (where

C

is the constant of integration)

Options

A $x + \ln \left|3sin x + cos x + 2\right| + c$
B $x - \ln \left|3sin x + cos x + 2\right| + c$
C $\ln \left|3sin x + cos x + 2\right| + c$
D $2x + \ln \left|3sin x + cos x + 2\right| + c$

Correct answer

C. $\ln \left|3sin x + cos x + 2\right| + c$

Step-by-step solution

$3J - I = ∫ \dfrac{3\cos x - \sin x}{3\sin x + \cos x + 2}dx$
Let $3\sin x + \cos x + 2 = t$
$⇒ \left(3cos x - sin x\right)dx = dt$
So, $3J - I = ∫ \dfrac{dt}{t} = \ln \left|t\right| + C$

Hence,

3J - I = \ln \left|3sin x + cos x + 2\right| + c

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