A) \[\int_{{}}^{{}}{{{x}^{6}}{{\tan }^{-1}}{{x}^{3}}}\ dx\]
B) \[\int_{{}}^{{}}{{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\ dx}\]
C) \[\int_{{}}^{{}}{{{x}^{3}}\cos {{x}^{2}}\ dx}\]
D) None of these
Correct Answer: C
Solution :
\[{{x}^{2}}=t\] is applicable for \[\int_{{}}^{{}}{{{x}^{3}}\cos {{x}^{2}}\,dx}\] \[=\frac{1}{2}\int_{{}}^{{}}{t\cos t\,dt}=\frac{1}{2}(t\sin t-\int_{{}}^{{}}{\sin t\,dt+c)}\] \[=\frac{1}{2}(t\sin t+\cos t+c)=\frac{1}{2}({{x}^{2}}\sin {{x}^{2}}+\cos {{x}^{2}}+c).\]
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