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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\ dx=}\]   [IIT 1986]

    A) \[\sin 2x+c\]

    B) \[-\frac{1}{2}\sin 2x+c\]

    C) \[\frac{1}{2}\sin 2x+c\]

    D) \[-\sin 2x+c\]

    Correct Answer: B

    Solution :

    • \[\int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\,dx}\]                   
    • \[=\int_{{}}^{{}}{\frac{({{\sin }^{4}}x+{{\cos }^{4}}x)({{\sin }^{4}}x-{{\cos }^{4}}x)}{{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x}}\,dx\]
    • \[=\int_{{}}^{{}}{({{\sin }^{4}}x-{{\cos }^{4}}x)\,dx}\]                   
    • \[=\int_{{}}^{{}}{({{\sin }^{2}}x+{{\cos }^{2}}x)({{\sin }^{2}}x-{{\cos }^{2}}x)\,dx}\]                
    • \[=\int_{{}}^{{}}{({{\sin }^{2}}x-{{\cos }^{2}}x)\,dx}\]\[=\int_{{}}^{{}}{-\cos 2x\,dx=-\frac{\sin 2x}{2}+c.}\]


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