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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main · Mathematics · Indefinite Integration

If ∫ 3 sin x + 2 cos x 3 cos x + 2 sin x dx = ax + b |2 sin x + 3...

If ∫ \dfrac{\text{3 sin x + 2 cos x}}{\text{3 cos x + 2 sin x}} \text{dx = ax + b} \text\ln \left|\text{2 sin x + 3 cos x}\right| + \text{C}, then

Options

  1. Aa=−1213, b =1539\text{a} = - \dfrac{\text{12}}{\text{13}} \text{, b =} \dfrac{\text{15}}{\text{39}}
  2. Ba=−713, b =613\text{a} = - \dfrac{\text{7}}{\text{13}} \text{, b =} \dfrac{\text{6}}{\text{13}}
  3. Ca=1213, b =−1539\text{a} = \dfrac{\text{12}}{\text{13}} \text{, b =} - \dfrac{\text{15}}{\text{39}}
  4. Da=−713, b =−613\text{a} = - \dfrac{\text{7}}{\text{13}} \text{, b =} - \dfrac{\text{6}}{\text{13}}

Correct answer

C. a=1213, b =−1539\text{a} = \dfrac{\text{12}}{\text{13}} \text{, b =} - \dfrac{\text{15}}{\text{39}}

Step-by-step solution

Differentiating both sides, we get

 3 sin x + 2 cos x3 cos x + 2 sin x==a+b(2 cos x - 3 sin x)(2 sin x + 3 cos x)~\dfrac{\text{3 sin x + 2 cos x}}{\text{3 cos x + 2 sin x}} = = \text{a} + \dfrac{\text{b(2 cos x - 3 sin x)}}{\text{(2 sin x + 3 cos x)}}

=sin x (2a - 3b) + cos x (3a + 2b)(3 cos x + 2 sin x)= \dfrac{\text{sin x (2a - 3b) + cos x (3a + 2b)}}{\left(\text{3 cos x + 2 sin x}\right)}

Comparing like terms on both sides, we get

3=2a−3b, 2 = 3a + 2b→a=1213, b =−15393 = 2 \text{a} - 3 \text{b, 2 = 3a + 2b} \to \text{a} = \dfrac{1 2}{1 3} \text{, b =} - \dfrac{1 5}{3 9}
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