Question

If $\int {\cos }^{4}xdx=Ax+B\sin 2x+C\sin 4x+D$, then $\{A,B,C\}$ equals

• A. $\{\frac{3}{8},\frac{1}{32},\frac{1}{4}\}$
• B. $\{\frac{3}{8},\frac{1}{4},\frac{1}{32}\}$
• C. $\{\frac{1}{32},\frac{1}{4},\frac{3}{8}\}$
• D. $\{\frac{1}{4},\frac{3}{8},\frac{1}{32}\}$

Answer 1

The correct option is B $\{\frac{3}{8},\frac{1}{4},\frac{1}{32}\}$

$\because \int {\cos }^{4}x=\frac{1}{4}\int (2{\cos }^{2}x{)}^{2}dx=\frac{1}{4}\int (1+\cos 2x{)}^{2}dx$
$=\frac{1}{4}\int (1+2\cos 2x+{\cos }^{2}2x)dx$
$=\frac{1}{8}\int (2+4\cos 2x+(1+\cos 4x\left)\right)dx$
$=\frac{1}{8}\{3x+2\sin 2x+\frac{1}{4}\sin 4x\}+D$
$=\frac{3}{8}x+\frac{1}{4}\sin 2x+\frac{1}{32}\sin 4x+D$

On comparing, we get
$A=\frac{3}{8},B=\frac{1}{4},C=\frac{1}{32}$
Hence, the correct option is (b).