Question

If $I=\displaystyle \int \frac{sin x}{3 sin ⁡ x + cos ⁡ x + 2}dx$ and $J=\displaystyle \int \frac{cos x}{3 sin ⁡ x + cos ⁡ x + 2}dx$ , then $3J-I$ is equal to (where $C$ is the constant of integration)

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

Answer 1

Answer: (C)

$3J-I=\displaystyle \int \frac{3 c o s x - s i n x}{3 s i n x + c o s x + 2}dx$
Let $3sinx+cosx+2=t$
$\Rightarrow \left(3 cos x - sin ⁡ x\right)dx=dt$
So, $3J-I=\displaystyle \int \frac{d t}{t}=ln \left|t\right|+C$
$3J-I=ln \left|3 sin ⁡ x + cos ⁡ x + 2\right|+c$