A) \[\frac{{{\cos }^{2}}x}{2}+c\]
B) \[\frac{{{\sin }^{2}}x}{3}+c\]
C) \[\frac{{{\sin }^{3}}x}{3}+c\]
D) \[-\frac{{{\cos }^{2}}x}{2}+c\]
Correct Answer: C
Solution :
\[I=\int_{{}}^{{}}{{{\sin }^{2}}x\,.\,\cos x\,dx}\]. Put \[\sin x=t\Rightarrow \cos x\,dx=dt\] \[\therefore \,\,\,I=\int_{{}}^{{}}{{{t}^{2}}dt}=\frac{{{t}^{3}}}{3}+c=\frac{{{\sin }^{3}}x}{3}+c\].You need to login to perform this action.
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