A) \[\frac{{{\sin }^{4}}x{{\cos }^{2}}x}{8}+c\]
B) \[\frac{{{\sin }^{4}}x}{4}+c\]
C) \[\frac{{{\sin }^{2}}x}{2}+c\]
D) \[4{{\sin }^{4}}x+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{{{\sin }^{3}}x\,.\,\cos x\,dx}\]. Put \[\sin x=t,\] then \[\cos x\,dx=dt\]; \[\int_{{}}^{{}}{{{t}^{3}}dt}=\frac{{{t}^{4}}}{4}=\frac{{{\sin }^{4}}x}{4}+c\].
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