Giải bài 3.45 trang 181 - SBT Đại số và Giải tích lớp 12
Tính các tích phân sau:
a) \(\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2x.{{\cos }^{2}}xdx}\)
b) \(\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{{{e}^{x}}}{{{e}^{2x}}-1}dx}\)
c) \(\int\limits_{0}^{1}{\dfrac{x+2}{{{x}^{2}}+2x+1}\ln(x+1)dx}\)
d) \(\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{x\sin x+\left( x+1 \right)\cos x}{x\sin x+\cos x}}dx\)
a) Vì \({{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}\) nên
\(\begin{aligned} & \int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2x.{{\cos }^{2}}xdx}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\cos 2x\left( 1+\cos 2x \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \cos 2x+{{\cos }^{2}}2x \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{4}}{\left( \cos 2x+\dfrac{1+\cos 4x}{2} \right)dx} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\left( \dfrac{1}{2}\sin 2x+\dfrac{1}{2}x+\dfrac{1}{4}\sin 4x \right)\left| \begin{aligned} & \dfrac{\pi }{4} \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\left( \dfrac{1}{2}+\dfrac{\pi }{8} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{4}+\dfrac{\pi }{16} \\ \end{aligned}\)
b) \(\dfrac{{{e}^{x}}}{{{e}^{2x}}-1}=\dfrac{1}{2}\left( \dfrac{{{e}^{x}}}{{{e}^{x}}-1}-\dfrac{{{e}^{x}}}{{{e}^{x}}+1} \right)\)
Ta có:
\( \begin{aligned} & \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{{{e}^{x}}}{{{e}^{2x}}-1}dx}=\dfrac{1}{2}\left( \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{{{e}^{x}}}{{{e}^{x}}-1}dx}-\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{{{e}^{x}}}{{{e}^{x}}+1}dx} \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\left[ \int\limits_{\dfrac{1}{2}}^{1}{\dfrac{d\left( {{e}^{x}}-1 \right)}{{{e}^{x}}-1}}-\int\limits_{\dfrac{1}{2}}^{1}{\dfrac{d\left( {{e}^{x}}+1 \right)}{{{e}^{x}}+1}} \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\ln \left| \dfrac{{{e}^{x}}-1}{{{e}^{x}}+1} \right|\left| \begin{aligned} & 1 \\ & \dfrac{1}{2} \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{2}\ln \dfrac{\left( e-1 \right)\left( \sqrt{e}+1 \right)}{\left( e+1 \right)\left( \sqrt{e}-1 \right)} \\ \end{aligned}\)
c)
\(\begin{aligned} & \int\limits_{0}^{1}{\frac{x+2}{{{x}^{2}}+2x+1}\ln (x+1)dx}=\int\limits_{0}^{1}{\frac{\left( x+1 \right)+1}{{{\left( x+1 \right)}^{2}}}\ln (x+1)dx} \\ & =\int\limits_{0}^{1}{\frac{\ln (x+1)}{x+1}dx}+\int\limits_{0}^{1}{\frac{\ln (x+1)}{{{\left( x+1 \right)}^{2}}}dx} \\ & =\frac{1}{2}{{\ln }^{2}}\left| x+1 \right|\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-\frac{\ln \left( x+1 \right)}{x+1}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-\frac{1}{x+1}\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & =\frac{1}{2}{{\ln }^{2}}2-\frac{1}{2}\ln 2+\frac{1}{2} \\ \end{aligned} \)
d)
\(\begin{aligned} & \int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{x\sin x+\left( x+1 \right)\cos x}{x\sin x+\cos x}}dx=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 1+\dfrac{x\cos x}{x\sin x+\cos x} \right)}dx \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\pi }{4}+\int\limits_{0}^{\dfrac{\pi }{4}}{\dfrac{d\left( x\sin x+\cos x \right)}{x\sin x+\cos x}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\pi }{4}+\ln \left| x\sin x+\cos x \right|\left| \begin{aligned} & \dfrac{\pi }{4} \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{\pi }{4}+\ln \left( 1+\dfrac{\pi }{4} \right)-\dfrac{1}{2}\ln 2 \\ \end{aligned}\)