Tính \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)
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Lời giải:
Báo saiTa có: \(\displaystyle \frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}\)\(\displaystyle = \frac{{\left( {x\sin x + \cos x} \right) + x\cos x}}{{x\sin x + \cos x}}\) \(\displaystyle = 1 + \frac{{x\cos x}}{{x\sin x + \cos x}}\)
Khi đó \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)\(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {\left( {1 + \frac{{x\cos x}}{{x\sin x + \cos x}}} \right)dx} \) \(\displaystyle = \int\limits_0^{\frac{\pi }{4}} {dx} + \int\limits_0^{\frac{\pi }{4}} {\frac{{x\cos x}}{{x\sin x + \cos x}}dx} \)
\(\displaystyle = \frac{\pi }{4} + I\) với \(\displaystyle I = \int\limits_0^{\frac{\pi }{4}} {\frac{{x\cos x}}{{x\sin x + \cos x}}dx} \)
Đặt \(\displaystyle x\sin x + \cos x = u\) \(\displaystyle \Rightarrow du = \left( {\sin x + x\cos x - \sin x} \right)dx\) \(\displaystyle = x\cos xdx\)
\(\displaystyle \Rightarrow I = \int\limits_1^{\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)} {\frac{{du}}{u}} \) \(\displaystyle = \left. {\ln \left| u \right|} \right|_1^{\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)}\) \(\displaystyle = \ln \left[ {\frac{{\sqrt 2 }}{2}\left( {\frac{\pi }{4} + 1} \right)} \right]\) \(\displaystyle = \ln \frac{{\sqrt 2 }}{2} + \ln \left( {\frac{\pi }{4} + 1} \right)\) \(\displaystyle = \ln \left( {1 + \frac{\pi }{4}} \right) - \frac{1}{2}\ln 2\)
Vậy \(\displaystyle \int\limits_0^{\frac{\pi }{4}} {\frac{{x\sin x + (x + 1)\cos x}}{{x\sin x + \cos x}}dx} \)\(\displaystyle = \frac{\pi }{4} + \ln \left( {1 + \frac{\pi }{4}} \right) - \frac{1}{2}\ln 2\).
