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Câu hỏi: Nếu $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx}=\dfrac{a}{b}\ln c$ thì $a+2b+3c$ bằng
A. $13$.
B. $14$.
C. $9$.
D. $11$.
Ta có $\sqrt{1+\sin 2x}=\sqrt{{{\sin }^{2}}x+2\sin x.\cos x+{{\cos }^{2}}x}=\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}=\left| \sin x+\cos x \right|$
Với mọi $x\in \left[ \dfrac{\pi }{4};\dfrac{\pi }{2} \right]$ ta có, $\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)>0$.
Do đó, $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\left| \sin x+\cos x \right|}dx=}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\sin x+\cos x}dx}$.
$=-\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\left( \sin x+\cos x \right)'}{\sin x+\cos x}dx}=\left. -\ln \left| \sin x+\cos x \right| \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}=-\ln 1+\ln \sqrt{2}=\dfrac{1}{2}\ln 2$.
Suy ra $a=1;b=2;c=2$. Vậy $a+2b+3c=11$.
A. $13$.
B. $14$.
C. $9$.
D. $11$.
Ta có $\sqrt{1+\sin 2x}=\sqrt{{{\sin }^{2}}x+2\sin x.\cos x+{{\cos }^{2}}x}=\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}=\left| \sin x+\cos x \right|$
Với mọi $x\in \left[ \dfrac{\pi }{4};\dfrac{\pi }{2} \right]$ ta có, $\sin x+\cos x=\sqrt{2}\sin \left( x+\dfrac{\pi }{4} \right)>0$.
Do đó, $I=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx}=\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\left| \sin x+\cos x \right|}dx=}\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{\sin x+\cos x}dx}$.
$=-\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\left( \sin x+\cos x \right)'}{\sin x+\cos x}dx}=\left. -\ln \left| \sin x+\cos x \right| \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}=-\ln 1+\ln \sqrt{2}=\dfrac{1}{2}\ln 2$.
Suy ra $a=1;b=2;c=2$. Vậy $a+2b+3c=11$.
Đáp án D.