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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\cos 2\theta \log \left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)\ d\theta =}\]       [IIT 1994]

    A) \[{{(\cos \theta -\sin \theta )}^{2}}\log \left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)\]

    B) \[{{(\cos \theta +\sin \theta )}^{2}}\log \left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)\]

    C) \[\frac{{{(\cos \theta -\sin \theta )}^{2}}}{2}\log \left( \frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta } \right)\]

    D) \[\frac{1}{2}\sin 2\theta \log \tan \left( \frac{\pi }{4}+\theta  \right)-\frac{1}{2}\log \sec 2\theta \]

    Correct Answer: D

    Solution :

    • We know that                                
    • \[\log \left( \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right)=\log \left( \frac{1+\tan \theta }{1-\tan \theta } \right)=\log \tan \left( \frac{\pi }{4}+\theta  \right)\]                                
    • \[\int_{{}}^{{}}{\sec \theta \,d\theta }=\log \tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right)\]                                
    • \[\therefore \,\,\,\int_{{}}^{{}}{\sec 2\theta \,d\theta }=\frac{1}{2}\log \tan \left( \frac{\pi }{4}+\theta  \right)\]                                
    • \[\therefore \,\,\,2\sec 2\theta =\frac{d}{d\theta }\log \tan \left( \frac{\pi }{4}+\theta  \right)\] .?.(i)                                
    • Integrating the given expression by parts, we get                                
    • \[I=\frac{1}{2}\sin 2\theta \log \tan \left( \frac{\pi }{4}+\theta  \right)-\frac{1}{2}\int_{{}}^{{}}{\sin 2\theta \,.\,2\sec 2\theta \,d\theta }\] by (i)                                  
    • \[=\frac{1}{2}\sin 2\theta \log \tan \left( \frac{\pi }{4}+\theta  \right)-\int_{{}}^{{}}{\tan 2\theta }\,d\theta \]                                                  
    • \[=\frac{1}{2}\sin 2\theta \log \tan \left( \frac{\pi }{4}+\theta  \right)-\frac{1}{2}\log \sec 2\theta \].


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