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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Mock Test - Integrals

  • question_answer
    If \[A=\int_{0}^{\pi }{\frac{\operatorname{cosx}}{{{(x+2)}^{2}}}dx,}\] then \[A=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx,}\] is equal to

    A) \[\frac{1}{2}+\frac{1}{\pi +2}-A\]

    B) \[\frac{1}{\pi +2}-A\]

    C) \[1+\frac{1}{\pi +2}-A\]

    D) \[A-\frac{1}{2}-\frac{1}{\pi +2}\]

    Correct Answer: A

    Solution :

    [a] \[I=\int_{0}^{\pi /2}{\frac{\sin 2x}{x+1}dx.}\] Put \[x=y/2.\]Then, \[I=\int_{0}^{\pi }{\frac{\sin y}{y+2}dy}\] \[=\left( \frac{-\cos y}{y+2} \right)_{0}^{\pi }-\int\limits_{0}^{\pi }{\frac{\cos y}{{{(y+2)}^{2}}}dy}\] (Integrating by parts) \[=\frac{1}{\pi +2}+\frac{1}{2}-A\]      


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