A) \[x+c\]
B) \[\frac{3}{2}\sin 2x+c\]
C) \[-\frac{3}{2}\cos 2x+c\]
D) \[\frac{1}{3}\sin 3x-\cos 3x+c\]
E) \[\frac{1}{3}\sin 3x+\cos 3x+c\]
Correct Answer: A
Solution :
Let \[I=\int{({{\sin }^{6}}x+{{\cos }^{6}}x+3{{\sin }^{2}}x{{\cos }^{2}}x)}dx\] \[=\int{\{{{({{\sin }^{2}}x)}^{3}}+{{({{\cos }^{2}}x)}^{3}}}\] \[+3{{\sin }^{2}}x{{\cos }^{2}}x\}dx\}\] \[=\int{\left[ \begin{align} & ({{\sin }^{2}}x+{{\cos }^{2}}x)({{\sin }^{4}}x+{{\cos }^{4}}x \\ & -{{\sin }^{2}}x{{\cos }^{2}}x)+3{{\sin }^{2}}x{{\cos }^{2}}x \\ \end{align} \right]}dx\] \[=\int{\left[ \begin{align} & {{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-3{{\sin }^{2}}x{{\cos }^{2}}x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3{{\sin }^{2}}x{{\cos }^{2}}x \\ \end{align} \right]}dx\] \[=\int{1\,dx\,\,x+c}\]You need to login to perform this action.
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