Class 12 Integrals · Long Answer3 Marksdefinite_integration
Class 12 Maths · Integrals · Long Answer #94
For real constants a,b evaluate \(\int_{-\pi}^{\pi}(\cos ax-\sin bx)^2dx\), including zero-parameter cases.
✅ Step-by-Step Solution
Final Answer: 2π+T(a)−T(b), where T(z)=sin(2πz)/(2z) for z≠0 and T(0)=π.
- Expand into cos²(ax)+sin²(bx)−2cos(ax)sin(bx). The cross term is odd and its symmetric integral is zero.
- Use cos²u=(1+cos2u)/2 and sin²u=(1−cos2u)/2. Their symmetric integrals are π+T(a) and π−T(b), with T(z)=sin(2πz)/(2z).
- At z=0 take T(0)=lim sin(2πz)/(2z)=π. Adding gives the stated formula, valid for all real a,b.
- Therefore the integral equals 2π+T(a)−T(b), where T(z)=sin(2πz)/(2z) for z≠0 and T(0)=π..
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] Expand into cos²(ax)+sin²(bx)−2cos(ax)sin(bx). The cross term is odd and its symmetric integral is zero.
- [1 mark] Use cos²u=(1+cos2u)/2 and sin²u=(1−cos2u)/2. Their symmetric integrals are π+T(a) and π−T(b), with T(z)=sin(2πz)/(2z).
- [1 mark] At z=0 take T(0)=lim sin(2πz)/(2z)=π. Adding gives the stated formula, valid for all real a,b.
- [0.5 marks] Therefore the integral equals 2π+T(a)−T(b), where T(z)=sin(2πz)/(2z) for z≠0 and T(0)=π..
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