A) \[-\frac{1}{4}{{\cos }^{4}}{{x}^{2}}+c\]
B) \[\frac{1}{4}{{\cos }^{4}}{{x}^{2}}+c\]
C) \[{{\cos }^{4}}{{x}^{2}}+c\]
D) None of these
Correct Answer: A
Solution :
Put \[t=\cos {{x}^{2}}\Rightarrow dt=-2x\sin {{x}^{2}}dx,\] then \[\int_{{}}^{{}}{2x{{\cos }^{3}}{{x}^{2}}\sin {{x}^{2}}dx}=-\int_{{}}^{{}}{{{t}^{3}}dt}=-\frac{{{t}^{4}}}{4}+c\] \[=-\frac{1}{4}{{\cos }^{4}}{{x}^{2}}+c.\]
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