Google
Câu hỏi: Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2}-x \right)dx}.$
A. $I=\dfrac{1-\sqrt{2}}{\sqrt{2}}$
B. $I=1-\sqrt{2}.$
C. $I=\dfrac{\sqrt{2}-1}{\sqrt{2}}.$
D. $I=\sqrt{2}-1.$
A. $I=\dfrac{1-\sqrt{2}}{\sqrt{2}}$
B. $I=1-\sqrt{2}.$
C. $I=\dfrac{\sqrt{2}-1}{\sqrt{2}}.$
D. $I=\sqrt{2}-1.$
Ta có
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\sin xdx}=-\cos x\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.=1-\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-1}{\sqrt{2}}.$
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\cos \left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{4}}{\sin xdx}=-\cos x\left| \begin{aligned}
& \dfrac{\pi }{4} \\
& 0 \\
\end{aligned} \right.=1-\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}-1}{\sqrt{2}}.$
Đáp án C.