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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{-\,\pi }^{\,\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}dx}\] is [AIEEE 2002]

    A) \[{{\pi }^{2}}/4\]                 

    B) \[{{\pi }^{2}}\]

    C) 0    

    D) \[\pi /2\]

    Correct Answer: B

    Solution :

    • \[I=\int_{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}dx}=\int_{-\pi }^{\pi }{\frac{2x}{1+{{\cos }^{2}}x}\,dx}+\int_{-\pi }^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}\,dx}\]           
    • Þ \[I=0+\int_{-\pi }^{\pi }{\frac{2x\sin x}{1+{{\cos }^{2}}x}\,dx}\]
    • \[\left[ \left. \begin{matrix}   
    • \int_{-a}^{a}{f(x)dx=2\int_{0}^{a}{f(x)\,dx},} & \text{if }f(-x)=f(x)  \\   \,\,=0, & \text{if }f(-x)=-f(x)  \\ \end{matrix} \right] \right.\]           
    • \[\Rightarrow I=2\int_{0}^{\pi }{\frac{2x\,\,\sin x}{1+{{\cos }^{2}}x}}\,dx\] \[\Rightarrow I=4\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}\,dx}\]-..(i)           
    • \[\Rightarrow I=4\int_{0}^{\pi }{\frac{(\pi -x)\,\,\sin x}{1+{{\cos }^{2}}x}}\,dx\]  --(ii)        
    • \[\left( \because \int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-x)\,dx}} \right)\]           
    • Adding (i) and (ii), we get           
    • \[\Rightarrow 2I=4\int_{0}^{\pi }{\frac{\pi \,\,\sin x}{1+{{\cos }^{2}}x}}\,dx\] \[\Rightarrow I=2\pi \int_{0}^{\pi }{\frac{\,\,\sin x}{1+{{\cos }^{2}}x}dx}\]           
    • Put \[\cos x=t\,\,\]Þ \[-\sin x\,\,dx=dt\]\[\Rightarrow I=2\pi \int_{1}^{-1}{\frac{-dt}{1+{{t}^{2}}}}\]           
    • \[\Rightarrow I=-2\pi \,[{{\tan }^{-1}}t]\,_{1}^{-1}\]\[\Rightarrow I=-2\pi \left( \frac{-\pi }{4}-\frac{\pi }{4} \right)={{\pi }^{2}}\].


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