A) \[\sin x-\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+c\]
B) \[\sin x+\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+c\]
C) \[\sin x-\frac{2}{3}{{\sin }^{3}}x-\frac{1}{5}{{\sin }^{5}}x+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{{{\cos }^{5}}x\,dx}=\int_{{}}^{{}}{{{\cos }^{4}}x\cos x\,dx}=\int_{{}}^{{}}{{{(1-{{\sin }^{2}}x)}^{2}}\cos xdx}\] Put \[\sin x=t\Rightarrow \cos x\ dx=dt\], then it reduces to \[\int_{{}}^{{}}{{{(1-{{t}^{2}})}^{2}}dt=\int_{{}}^{{}}{(1+{{t}^{4}}-2{{t}^{2}})\ dt=\frac{{{t}^{5}}}{5}-\frac{2{{t}^{3}}}{3}+t+c}}\] \[=\frac{{{\sin }^{5}}x}{5}-\frac{2{{\sin }^{3}}x}{3}+\sin x+c\].
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