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  5. Definite Integrals

12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    The value of \[\int_{\,-\,\pi }^{\,\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}dx,\,a>0,}\] is [IIT Screening 2001; AIEEE 2005]

    A) \[\pi \]                                    

    B) \[a\pi \]

    C) \[\frac{\pi }{2}\]                 

    D) \[2\pi \]

    Correct Answer: C

    Solution :

    • \[I=\int_{-\pi }^{\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{x}}}dx}=\int_{\,\pi }^{\,-\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}(-dx)}\]  (Putting? x for x)             
    • \[=\int_{\,-\pi }^{\,\pi }{\frac{{{\cos }^{2}}x}{1+{{a}^{-x}}}}\,dx\]\[\Rightarrow I+I=\int_{\,-\pi }^{\,\pi }{{{\cos }^{2}}x\left( \frac{1}{1+{{a}^{x}}}+\frac{1}{1+{{a}^{-x}}} \right)\,dx}\]          
    • \[=\int_{\,-\pi }^{\,\pi }{{{\cos }^{2}}x\,dx}\]Þ 2I \[=2\int_{0}^{\pi }{{{\cos }^{2}}x.\,dx=}\int_{0}^{\pi }{(1+\cos 2x)dx}\]           
    • Þ 2I \[=[x]_{0}^{\pi }+\left[ \frac{\sin 2x}{2} \right]_{0}^{\pi }\]\[\Rightarrow 2I=\pi \,\,\,\Rightarrow I=\frac{\pi }{2}\].


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