Google
Câu hỏi: Tìm nguyên hàm $F\left( x \right)$ của hàm số $f\left( x \right)=\cos x\sqrt{\sin x+1}.$
A. $F\left( x \right)=\dfrac{1}{3}\sin x\sqrt{\sin x+1}+C.$
B. $F\left( x \right)=\dfrac{1-2\sin x-3{{\sin }^{2}}x}{2\sqrt{\sin x+1}}.$
C. $F\left( x \right)=\dfrac{1}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$.
D. $F\left( x \right)=\dfrac{2}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$.
A. $F\left( x \right)=\dfrac{1}{3}\sin x\sqrt{\sin x+1}+C.$
B. $F\left( x \right)=\dfrac{1-2\sin x-3{{\sin }^{2}}x}{2\sqrt{\sin x+1}}.$
C. $F\left( x \right)=\dfrac{1}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$.
D. $F\left( x \right)=\dfrac{2}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$.
$I=F\left( x \right)=\int\limits_{{}}^{{}}{\cos x\sqrt{\sin x+1}dx}$
Đặt $u=\sqrt{\sin x+1}\Rightarrow {{u}^{2}}=\sin x+1$
$\Rightarrow 2udu=\cos xdx.$
$I=\int\limits_{{}}^{{}}{u.2udu}=2\int\limits_{{}}^{{}}{{{u}^{2}}du}$
$=\dfrac{2}{3}{{u}^{3}}+C=\dfrac{2}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$
Đặt $u=\sqrt{\sin x+1}\Rightarrow {{u}^{2}}=\sin x+1$
$\Rightarrow 2udu=\cos xdx.$
$I=\int\limits_{{}}^{{}}{u.2udu}=2\int\limits_{{}}^{{}}{{{u}^{2}}du}$
$=\dfrac{2}{3}{{u}^{3}}+C=\dfrac{2}{3}\left( \sin x+1 \right)\sqrt{\sin x+1}+C$
Đáp án D.