A) \[{{\sin }^{3}}x+c\]
B) \[\frac{\sin 3x}{3}+\frac{\sin x}{2}+c\]
C) \[\frac{\sin 3x}{12}+\frac{3}{4}\sin x+c\]
D) \[\frac{\sin 3x}{4}+3\sin x+c\]
Correct Answer: C
Solution :
Let \[I=\int{{{\cos }^{3}}x}dx\] \[\Rightarrow \] \[I=\int{{{\cos }^{2}}x.\cos xdx}\] \[\Rightarrow \] \[I=\int{(1-{{\sin }^{2}}x)}\cos xdx\] Let \[sin\text{ }x=t\] \[\Rightarrow \] \[cos\text{ }xdx=dt\] \[\therefore \] \[I=\int{(1-{{t}^{2}})}dt\] \[=t-\frac{{{t}^{3}}}{3}+c\] \[=\sin x-\frac{{{\sin }^{3}}x}{3}+c\] \[=\sin x-\frac{1}{3}\left[ \frac{1}{4}(3\sin x-\sin 3x) \right]+c\] \[[\because \sin 3x=3\sin x-4{{\sin }^{3}}x]\] \[=\sin x-\frac{1}{4}\sin x+\frac{\sin 3x}{12}+c\] \[=\frac{\sin 3x}{12}+\frac{3}{4}\sin x+c\]
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