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Câu hỏi: Nguyên hàm $\int{\left[ {{\sin }^{2}}\left( 3x+1 \right)+\cos x \right]}dx$ là:
A. $\dfrac{1}{2}x-3\sin \left( 6x+2 \right)+\sin x+C$.
B. $x-3\sin \left( 6x+2 \right)+\sin x+C$.
C. $\dfrac{1}{2}x-3\sin \left( 3x+1 \right)+\sin x+C$.
D. $\dfrac{1}{2}x-3\sin \left( 6x+2 \right)-\sin x+C$.
A. $\dfrac{1}{2}x-3\sin \left( 6x+2 \right)+\sin x+C$.
B. $x-3\sin \left( 6x+2 \right)+\sin x+C$.
C. $\dfrac{1}{2}x-3\sin \left( 3x+1 \right)+\sin x+C$.
D. $\dfrac{1}{2}x-3\sin \left( 6x+2 \right)-\sin x+C$.
$\begin{aligned}
& \int{\left[ {{\sin }^{2}}\left( 3x+1 \right)+\cos x \right]}dx=\int{\left[ \dfrac{1-\cos \left( 6x+2 \right)}{2}+\cos x \right]}dx=\int{\left[ \dfrac{1}{2}-\dfrac{1}{2}\cos \left( 6x+2 \right)+\cos x \right]}dx \\
& =\dfrac{1}{2}x-3\sin \left( 6x+2 \right)+\sin x+C \\
\end{aligned}$
& \int{\left[ {{\sin }^{2}}\left( 3x+1 \right)+\cos x \right]}dx=\int{\left[ \dfrac{1-\cos \left( 6x+2 \right)}{2}+\cos x \right]}dx=\int{\left[ \dfrac{1}{2}-\dfrac{1}{2}\cos \left( 6x+2 \right)+\cos x \right]}dx \\
& =\dfrac{1}{2}x-3\sin \left( 6x+2 \right)+\sin x+C \\
\end{aligned}$
Đáp án A.