A) \[\left\{ \frac{3}{8},\,\frac{1}{32},\,\frac{1}{4} \right\}\]
B) \[\left\{ \frac{3}{8},\,\frac{1}{4},\,\frac{1}{32} \right\}\]
C) \[\left\{ \frac{1}{32},\,\frac{1}{4},\,\frac{3}{8} \right\}\]
D) \[\left\{ \frac{1}{4},\,\frac{3}{8},\,\frac{1}{32} \right\}\]
Correct Answer: B
Solution :
\[\because \] \[\int_{{}}^{{}}{{{\cos }^{4}}x\,dx=\frac{1}{4}}\,\int_{{}}^{{}}{{{(2\,{{\cos }^{2}}x)}^{2}}dx}\] \[=\frac{1}{4}\,\int{{{(1+\cos \,2x)}^{2}}dx}\] \[=\frac{1}{4}\,\int{(1+\cos \,2x+{{\cos }^{2}}2x)\,dx}\] \[=\frac{1}{8}\,\int{(2+4\,\cos \,2x+(1+\cos \,4x))dx}\] \[=\frac{1}{8}\,\left\{ 3x+2\,\sin \,2x+\frac{1}{4}\sin \,4x \right\}+D\] \[=\frac{3}{8}x+\frac{1}{4}\sin \,2x+\frac{1}{32}\,\sin \,4x+D\] On comparing the given integral, we get \[A=\frac{3}{8},\,B=\frac{1}{4},\,C=\frac{1}{32}\]
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