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Class 12 Integrals · Long Answer3 Marksindefinite_integration

Class 12 Maths · Integrals · Long Answer #78

Evaluate \(\int \frac{1}{\sin{\left(x \right)} + \sin{\left(2 x \right)}}\,dx\). Work on each connected open interval where the real integrand is defined.

✅ Step-by-Step Solution

Final Answer: \(\frac{\log{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{6} + \frac{\log{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \frac{2 \log{\left(\left|{2 \cos{\left(x \right)} + 1}\right| \right)}}{3}\) + C
  1. Factor the denominator as sin x(1+2cos x). Multiply numerator and denominator by sin x and set u=cos x, du=−sin x dx.
  2. The transformed integrand is −1/[(1−u²)(1+2u)]=1/[2(1+u)]−1/[6(1−u)]−4/[3(1+2u)].
  3. Integrate to (1/2)ln|1+u|+(1/6)ln|1−u|−(2/3)ln|1+2u| and replace u by cos x.
  4. Therefore the required integral is \(\frac{\log{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{6} + \frac{\log{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \frac{2 \log{\left(\left|{2 \cos{\left(x \right)} + 1}\right| \right)}}{3}\) + C.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Factor the denominator as sin x(1+2cos x). Multiply numerator and denominator by sin x and set u=cos x, du=−sin x dx.
  • [1 mark] The transformed integrand is −1/[(1−u²)(1+2u)]=1/[2(1+u)]−1/[6(1−u)]−4/[3(1+2u)].
  • [1 mark] Integrate to (1/2)ln|1+u|+(1/6)ln|1−u|−(2/3)ln|1+2u| and replace u by cos x.
  • [0.5 marks] Therefore the required integral is \(\frac{\log{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{6} + \frac{\log{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} - \frac{2 \log{\left(\left|{2 \cos{\left(x \right)} + 1}\right| \right)}}{3}\) + C.

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