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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{(\sin x+\sin 2x)}=}\]       [IIT 1984]

    A) \[\frac{1}{6}\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)-\frac{2}{3}\log (1+2\cos x)\]        

    B) \[6\log (1-\cos x)+2\log (1+\cos x)-\frac{2}{3}\log (1+2\cos x)\]

    C) \[6\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)+\frac{2}{3}\log (1+2\cos x)\]       

    D) None of these

    Correct Answer: A

    Solution :

    • \[I=\int_{{}}^{{}}{\frac{dx}{\sin x(1+2\cos x)}}=\int_{{}}^{{}}{\frac{\sin x\,dx}{{{\sin }^{2}}x(1+2\cos x)}}\]                   
    • \[=\int_{{}}^{{}}{\frac{\sin x\,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}}\]                   
    • Now differential coefficient of \[\cos x\] is \[-\sin x\] which is given in numerator and hence we make the substitution \[\cos x=t\Rightarrow -\sin x\,dx=dt\]                   
    • \[\therefore \,\,\,I=-\int_{{}}^{{}}{\frac{dt}{(1-t)(1+t)(1+2t)}}\]                   
    • We split the integrand into partial fractions                
    • \ \[I=-\int{\left[ \frac{1}{6(1-t)}-\frac{1}{2(1+t)}+\frac{4}{3(1+2t)} \right]}\,dt\] etc.


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