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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int\limits_{0}^{\pi }{\,\frac{\sin \left( n+\frac{1}{2} \right)\text{ }x}{\sin x}}\,dx\], \[(n\in N)\] equals [Kurukshetra CEE 1998]

    A) \[n\pi \]                                 

    B) \[(2n+1)\frac{\pi }{2}\]

    C) \[\pi \]                                    

    D) 0

    Correct Answer: C

    Solution :

    • \[2\sin \frac{x}{2}.\left( \frac{1}{2}+\cos x+\cos 2x+.....+\cos nx \right)\]                   
    • \[=\sin \frac{x}{2}+2\sin \frac{x}{2}\cos x+2\sin \frac{x}{2}\cos 2x+....+2\sin \frac{x}{2}\cos nx\]           
    • \[=\sin \frac{x}{2}+\sin \frac{3x}{2}-\sin \frac{x}{2}+\sin \frac{5x}{2}-\sin \frac{3x}{2}+.....\]                              
    • \[+\sin \left( n+\frac{1}{2} \right)x-\sin \left( n-\frac{1}{2} \right)x\]\[=\sin \left( n+\frac{1}{2} \right)x\]                   
    • \ \[\frac{1}{2}+\cos x+\cos 2x+.....+\cos nx=\frac{\sin \left( n+\frac{1}{2} \right)x}{2\sin \left( \frac{x}{2} \right)}\]                   
    • Þ \[\int_{0}^{\pi }{\frac{\sin \left( n+\frac{1}{2} \right)x}{\sin \left( \frac{x}{2} \right)}dx}=2\left( \int_{0}^{\pi }{\frac{1}{2}dx+\int_{0}^{\pi }{\cos xdx+.....+\int_{0}^{\pi }{\cos nx\,dx}}} \right)\]                                                                        
    • \[=2\left( \frac{\pi }{2}+\sin x+.....+\frac{\sin nx}{n} \right)_{0}^{\pi }=\pi \].


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