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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{2}}}{{{(x\sin x+\cos x)}^{2}}}\ dx=}\]          [MNR 1989; RPET 2000]

    A) \[\frac{\sin x+\cos x}{x\sin x+\cos x}\]

    B) \[\frac{x\sin x-\cos x}{x\sin x+\cos x}\]

    C) \[\frac{\sin x-x\cos x}{x\sin x+\cos x}\]

    D) None of these

    Correct Answer: C

    Solution :

    • Differentiation of\[x\sin x+\cos x\]is \[x\cos x,\] then                    
    • \[I=\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(x\sin x+\cos x)}^{2}}}}=\int_{{}}^{{}}{\frac{x\cos x}{{{(x\sin x+\cos x)}^{2}}}.\frac{x}{\cos x}dx}\]                   
    • Integrate by parts \[\left[ \int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt=-\frac{1}{t}} \right]\]                   
    • \[\therefore \,\,\,I=\frac{-1}{(x\sin x+\cos x)}.\frac{x}{\cos x}\]                   
    • \[+\int_{{}}^{{}}{\frac{1}{(x\sin x+\cos x)}}.\frac{\cos x\,.\,1-x(-\sin x)}{{{\cos }^{2}}x}\,dx\]                          
    • \[=-\frac{1}{x\sin x+\cos x}.\frac{x}{\cos x}+\int_{{}}^{{}}{{{\sec }^{2}}x\,dx}\]                         
    • \[=-\frac{1}{x\sin x+\cos x}.\frac{x}{\cos x}+\frac{\sin x}{\cos x}\]                         
    • \[=\frac{-x+x{{\sin }^{2}}x+\sin x\cos x}{(x\sin x+\cos x)\cos x}\]                      
    • \[=\frac{\sin x\cos x-x(1-{{\sin }^{2}}x)}{(x\sin x+\cos x)\cos x}\]\[=\frac{\sin x-x\cos x}{x\sin x+\cos x}\].


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