Giải bài 3.43 trang 180 - SBT Đại số và Giải tích lớp 12
Tính các nguyên hàm sau:
a) \(\int{\left( 2x-3 \right)\sqrt{x-3}dx}\) , đặt \(u=\sqrt{x-3}\)
b) \(\int{\dfrac{x}{{{\left( 1+{{x}^{2}} \right)}^{\frac{3}{2}}}}dx}\), đặt \(u=\sqrt{{{x}^{2}}+1}\)
c) \(\int{\dfrac{{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}}dx}\) , đặt \(u={{e}^{2x}}+1\)
d) \(\int{\dfrac{1}{\sin x-\sin \alpha }dx}\)
a) Đặt \(u=\sqrt{x-3}\Rightarrow {{u}^{2}}=x-3\Rightarrow 2udu=dx\)
Ta có:
\(\begin{aligned} & \int{\left( 2x-3 \right)\sqrt{x-3}dx}=\int{\left[ 2\left( {{u}^{2}}+3 \right)-3 \right].u.2udu} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int{\left( 4{{u}^{4}}+6{{u}^{2}} \right)du} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{4}{5}{{u}^{5}}+2{{u}^{3}}+C \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{4}{5}{{\left( x-3 \right)}^{\frac{5}{2}}}+2{{\left( x-3 \right)}^{\frac{3}{2}}}+C \\ \end{aligned}\)
b) Đặt \(u=\sqrt{{{x}^{2}}+1}\Rightarrow {{u}^{2}}={{x}^{2}}+1\Rightarrow udu=xdx\)
Ta có:
c)
d) Ta có: \(\cos a=\cos \left( \dfrac{x-a}{2}-\dfrac{x+a}{2} \right)\)
Suy ra:
\(\begin{aligned} & \frac{1}{\sin x-\sin a}=\dfrac{1}{\cos a}\dfrac{\cos \dfrac{x-a}{2}\cos \dfrac{x+a}{2}+\sin \dfrac{x-a}{2}\sin \dfrac{x+a}{2}}{2\cos \dfrac{x+a}{2}\sin \dfrac{x-a}{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{2}{\cos a}\left( \dfrac{\cos \dfrac{x-a}{2}}{\sin \dfrac{x-a}{2}}+\dfrac{\sin \dfrac{x+a}{2}}{\cos \dfrac{x+a}{2}} \right) \\ \end{aligned}\)
Do đó:
\(\begin{aligned} & \int{\dfrac{2}{\cos a}\left( \dfrac{\cos \dfrac{x-a}{2}}{\sin \dfrac{x-a}{2}}+\dfrac{\sin \dfrac{x+a}{2}}{\cos \dfrac{x+a}{2}} \right)dx}=\dfrac{1}{\cos a}\left( \ln \left| \sin \dfrac{x-a}{2} \right|-\ln \left| \cos \dfrac{x+a}{2} \right| \right)+C \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{1}{\cos a}\ln \left| \dfrac{\sin \dfrac{x-a}{2}}{\cos \dfrac{x+a}{2}} \right|+C \\ \end{aligned}\)