A) \[-\frac{{{\sin }^{4}}x}{4}+c\]
B) \[-\frac{{{\cos }^{4}}x}{4}+c\]
C) \[\frac{{{e}^{\sin }}x}{4}+c\]
D) None of these
Correct Answer: B
Solution :
Let\[I=\int_{{}}^{{}}{{{\cos }^{3}}x{{e}^{\log \sin x}}}dx\] \[=\int_{{}}^{{}}{{{\cos }^{3}}x}\sin xdx\] Put\[\cos x=t\Rightarrow -\sin xdx=dt\] \[\therefore \]\[I=-\int_{{}}^{{}}{{{t}^{3}}dt}\] \[=-\frac{{{t}^{4}}}{4}+c=-\frac{{{\cos }^{4}}x}{4}+c\]
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