Class 12 Integrals · Long Answer3 Marksindefinite_integration
Class 12 Maths · Integrals · Long Answer #68
Evaluate \(\int \frac{\cos{\left(x \right)}}{\sin{\left(3 x \right)}}\,dx\). Work on intervals with sin3x≠0.
✅ Step-by-Step Solution
Final Answer: \(- \frac{\log{\left(\left|{4 \sin^{2}{\left(x \right)} - 3}\right| \right)}}{6} + \frac{\log{\left(\left|{\sin{\left(x \right)}}\right| \right)}}{3}\) + C
- Set u=sin x, du=cos x dx, and sin3x=3u−4u³.
- Decompose 1/[u(3−4u²)]=1/(3u)+(4u)/(3(3−4u²)).
- Integrating gives (1/3)ln|u|−(1/6)ln|3−4u²|. Substitute u=sin x.
- Therefore the required integral is \(- \frac{\log{\left(\left|{4 \sin^{2}{\left(x \right)} - 3}\right| \right)}}{6} + \frac{\log{\left(\left|{\sin{\left(x \right)}}\right| \right)}}{3}\) + C.
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Set u=sin x, du=cos x dx, and sin3x=3u−4u³.
- [1 mark] Decompose 1/[u(3−4u²)]=1/(3u)+(4u)/(3(3−4u²)).
- [1 mark] Integrating gives (1/3)ln|u|−(1/6)ln|3−4u²|. Substitute u=sin x.
- [0.5 marks] Therefore the required integral is \(- \frac{\log{\left(\left|{4 \sin^{2}{\left(x \right)} - 3}\right| \right)}}{6} + \frac{\log{\left(\left|{\sin{\left(x \right)}}\right| \right)}}{3}\) + C.
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