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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{{{e}^{x/2}}\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)\ dx=}\]    [Roorkee 1982]

    A) \[{{e}^{x/2}}\cos \frac{x}{2}+c\]

    B) \[\sqrt{2}{{e}^{x/2}}\cos \frac{x}{2}+c\]

    C) \[{{e}^{x/2}}\sin \frac{x}{2}+c\]

    D) \[\sqrt{2}{{e}^{x/2}}\sin \frac{x}{2}+c\]

    Correct Answer: D

    Solution :

    • Let \[I=\int_{{}}^{{}}{{{e}^{x/2}}\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)dx}\]                   
    • \[=2\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)\,{{e}^{x/2}}-\int_{{}}^{{}}{\cos \left( \frac{x}{2}+\frac{\pi }{4} \right)\frac{1}{2}2{{e}^{x/2}}dx+c}\]                    \[=2\sin \left( \frac{x}{2}+\frac{\pi }{4} \right){{e}^{x/2}}-2{{e}^{x/2}}\cos \,\left( \frac{x}{2}+\frac{\pi }{4} \right)-\int_{{}}^{{}}{\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)\frac{1}{2}2{{e}^{x/2}}}\]                   
    • Therefore, \[2I=2{{e}^{x/2}}\left\{ \sin \left( \frac{x}{2}+\frac{\pi }{4} \right)-\cos \left( \frac{x}{2}+\frac{\pi }{4} \right) \right\}\]                   
    • \[\Rightarrow I={{e}^{x/2}}\left\{ \sin \left( \frac{x}{2}+\frac{\pi }{4} \right)-\cos \left( \frac{x}{2}+\frac{\pi }{4} \right) \right\}\]                          
    • \[=\sqrt{2}{{e}^{x/2}}\left( \sin \frac{x}{2} \right)=\sqrt{2}{{e}^{x/2}}\sin \frac{x}{2}+c.\]                   
    • Trick : By inspection,                   
    • \[\frac{d}{dx}\left\{ \sqrt{2}{{e}^{x/2}}\sin \frac{x}{2}+c \right\}=\sqrt{2}\left[ \frac{1}{2}{{e}^{x/2}}\cos \frac{x}{2}+\frac{1}{2}{{e}^{x/2}}\sin \frac{x}{2} \right]\]
    • \[={{e}^{x/2}}\left[ \frac{1}{\sqrt{2}}\cos \frac{x}{2}+\frac{1}{\sqrt{2}}\sin \frac{x}{2} \right]={{e}^{x/2}}\sin \left( \frac{x}{2}+\frac{\pi }{4} \right)\].


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