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Câu hỏi: Một nguyên hàm của hàm số $f\left( x \right)={{x}^{2}}\sin \left( {{x}^{3}} \right)$ là:
A. $3\cos \left( {{x}^{3}} \right).$
B. $\dfrac{1}{3}\sin \left( {{x}^{3}} \right).$
C. $-\dfrac{1}{3}\cos \left( {{x}^{3}} \right).$
D. $\dfrac{1}{3}\cos \left( {{x}^{3}} \right).$
A. $3\cos \left( {{x}^{3}} \right).$
B. $\dfrac{1}{3}\sin \left( {{x}^{3}} \right).$
C. $-\dfrac{1}{3}\cos \left( {{x}^{3}} \right).$
D. $\dfrac{1}{3}\cos \left( {{x}^{3}} \right).$
Cách 1: Ta tính $I=\int{f\left( x \right)dx}=\int{{{x}^{2}}\sin \left( {{x}^{3}} \right)dx}.$
Đặt $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx\Rightarrow \dfrac{1}{3}dt={{x}^{2}}dx.$
$I=\dfrac{1}{3}\int{\sin tdt}=-\dfrac{1}{3}\cos t+C=-\dfrac{1}{3}\cos \left( {{x}^{3}} \right)+C.$
Cách 2: $\int{{{x}^{2}}\sin \left( {{x}^{3}} \right)dx}=\dfrac{1}{3}\int{\sin \left( {{x}^{3}} \right)d\left( {{x}^{3}} \right)}=-\dfrac{1}{3}\cos \left( {{x}^{3}} \right)+C$
Đặt $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx\Rightarrow \dfrac{1}{3}dt={{x}^{2}}dx.$
$I=\dfrac{1}{3}\int{\sin tdt}=-\dfrac{1}{3}\cos t+C=-\dfrac{1}{3}\cos \left( {{x}^{3}} \right)+C.$
Cách 2: $\int{{{x}^{2}}\sin \left( {{x}^{3}} \right)dx}=\dfrac{1}{3}\int{\sin \left( {{x}^{3}} \right)d\left( {{x}^{3}} \right)}=-\dfrac{1}{3}\cos \left( {{x}^{3}} \right)+C$
Đáp án C.