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

  • question_answer
    \[\int \frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx\] is equal to

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

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

    C) \[-\frac{1}{2}\sin x+C\] 

    D) \[-{{\sin }^{2}}x+C\]

    Correct Answer: B

    Solution :

    [b] \[\int{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}\] \[=\int{\frac{(si{{n}^{2}}x-co{{s}^{2}}x)(si{{n}^{4}}x+co{{s}^{4}}x)}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}}\] \[=\int{-\cos 2xdx=-\frac{1}{2}\sin 2x+C}\]


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