# Bài 3 trang 113 SGK Giải tích 12

##### Hướng dẫn giải

a) Đặt $$u= x+1$$.

b) Đặt $$x = sint$$.

c) Đặt $$u = 1 + x.{e^x}$$.

d) Đặt $$x= asint$$.

Lời giải chi tiết

a) Đặt $$u= x+1 \Rightarrow du = dx$$ và $$x = u - 1$$.

Đổi cận: $$\left\{ \begin{array}{l}x = 0 \Rightarrow u = 1\\x = 3 \Rightarrow u = 4\end{array} \right.$$

$$\begin{array}{l}\int\limits_0^3 {\frac{{{x^2}}}{{{{\left( {1 + x} \right)}^{\frac{3}{2}}}}}dx} = \int\limits_1^4 {\frac{{{{\left( {u - 1} \right)}^2}}}{{{u^{\frac{3}{2}}}}}du} = \int\limits_1^4 {\frac{{{u^2} - 2u + 1}}{{{u^{\frac{3}{2}}}}}du} \\= \int\limits_1^4 {\left( {{u^{\frac{1}{2}}} - 2{u^{ - \frac{1}{2}}} + {u^{ - \frac{3}{2}}}} \right)du} \\= \left. {\left( {\frac{2}{3}{u^{\frac{3}{2}}} - 4{u^{\frac{1}{2}}} - 2{u^{ - \frac{1}{2}}}} \right)} \right|_1^4\\= - \frac{{11}}{3} - \left( { - \frac{{16}}{3}} \right) = \frac{5}{3}\end{array}$$

b) Đặt $$x = sint$$, $$0<t<\frac{\pi}{2}$$. Ta có: $$dx = costdt$$

và $$\sqrt{1-x^{2}}=\sqrt{1-sin^{2}t}= \sqrt{cos^{2}t}=\left | cost \right |= cos t.$$

Đổi cận: $$\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = 1 \Rightarrow t = \frac{\pi }{2}\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {\sqrt {1 - {x^2}} dx} = \int\limits_0^{\frac{\pi }{2}} {\sqrt {1 - {{\sin }^2}t} \cos tdt} \\= \int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} = \frac{1}{2}\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)dt} \\= \frac{1}{2}\left. {\left( {t + \frac{{\sin 2t}}{2}} \right)} \right|_0^{\frac{\pi }{2}}\\= \frac{1}{2}.\frac{\pi }{2} = \frac{\pi }{4}\end{array}$$

c) Đặt: $$t = 1 + x.{e^x} \Rightarrow dt = \left( {{e^x} + x.{e^x}} \right)dx = {e^x}\left( {1 + x} \right)dx$$.

Đổi cận: $$\left\{ \begin{array}{l}x = 0 \Rightarrow t = 1\\x = 1 \Rightarrow t = 1 + e\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {\frac{{{e^x}\left( {1 + x} \right)}}{{1 + x{e^x}}}dx} = \int\limits_1^{1 + e} {\frac{{dt}}{t}} = \left. {\ln \left| t \right|} \right|_1^{1 + e}\\= \ln \left( {1 + e} \right) - \ln 1 = \ln \left( {1 + e} \right)\end{array}$$

d) Đặt $$x = a\sin t \Rightarrow dx = a\cos tdt$$

Đổi cận: $$\left\{ \begin{array}{l}x = 0 \Rightarrow t = 0\\x = \frac{a}{2} \Rightarrow t = \frac{\pi }{6}\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^{\frac{a}{2}} {\frac{1}{{\sqrt {{a^2} - {x^2}} }}dx} = \int\limits_0^{\frac{\pi }{6}} {\frac{{a\cos tdt}}{{\sqrt {{a^2} - {a^2}{{\sin }^2}t} }}} \\= \int\limits_0^{\frac{\pi }{6}} {\frac{{a\cos tdt}}{{a.\cos t}}} = \int\limits_0^{\frac{\pi }{6}} {dt} = \left. t \right|_0^{\frac{\pi }{6}} = \frac{\pi }{6}\end{array}$$.