Bài 1 trang 112 - SGK Giải tích 12
Tính các tích phân sau:
a)\(\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx\) b) \(\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx\)
c)\(\int_{\frac{1}{2}}^{2}\frac{1}{x(x+1)}dx\) d) \(\int_{0}^{2}x(x+1)^{2}dx\)
e)\(\int_{\frac{1}{2}}^{2}\frac{1-3x}{(x+1)^{2}}dx\) g) \(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin3xcos5xdx\)
Giải
a) \(\int_{\frac{-1}{2}}^{\frac{1}{2}}\sqrt[3]{ (1-x)^{2}}dx\) = \(-\int_{\frac{-1}{2}}^{\frac{1}{2}}(1-x)^{\frac{2}{3}}d(1-x)=-\frac{3}{5}(1-x)^{\frac{5}{3}}|_{\frac{-1}{2}}^{\frac{1}{2}}\)
= \(-\frac{3}{5}\left [ \frac{1}{2\sqrt[3]{4}}-\frac{3\sqrt[3]{9}}{2\sqrt[3]{4}} \right ]=\frac{3}{10\sqrt[3]{4}}(3\sqrt[3]{9}-1)\)
b) \(\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)dx\)=\(-\int_{0}^{\frac{\pi}{2}}sin(\frac{\pi}{4}-x)d(\frac{\pi}{4}-x)\) = \(cos(\frac{\pi}{4}-x)|_{0}^{\frac{\pi}{2}}\)
= \(cos(\frac{\pi}{4}-\frac{\pi}{2})-cos\frac{\pi}{4}=0\)
c)\(\int_{\frac{1}{2}}^{2}\frac{1}{x(x+1)}dx\)=\(\int_{\frac{1}{2}}^{2}(\frac{1}{x}-\frac{1}{x+1})dx =ln\left | \frac{x}{x+1} \right ||_{\frac{1}{2}}^{2}=ln2\)
d)\(\int_{0}^{2}x(x+1)^{2}dx\)= \(\int_{0}^{2}(x^{3}+2x^{2}+x)dx=(\frac{x^{4}}{4}+\frac{2}{3}x^{3}+\frac{x^{2}}{2})|_{0}^{2}\)
= \(\frac{16}{4}+\frac{16}{3}+2= 11\tfrac{1}{3}\)
e)\(\int_{\frac{1}{2}}^{2}\frac{1-3x}{(x+1)^{2}}dx\)= \(\int_{\frac{1}{2}}^{2}\frac{-3(x+1)+4}{(x+1)^{2}}dx=\int_{\frac{1}{2}}^{2}\left [ \frac{-3}{x+1}+\frac{4}{(x+1)^{2}} \right ]dx\)
= \(\left ( -3.ln\left | x+1 \right |-\frac{4}{x+1} \right )|_{\frac{1}{2}}^{2}= \frac{4}{3}-3ln2\)
g)Ta có \(f(x) = sin3xcos5x\) là hàm số lẻ.
Vì \(f(-x) = sin(-3x)cos(-5x)\)
\(= -sin3xcos5x = -f(x)\)
nên:
\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin3xcos5x =0\)
Chú ý: Có thể tính trực tiếp bằng cách đặt \(x= -t\) hoặc biến đổi thành tổng.
Bài 2 trang 112 - SGK Giải tích 12
Tính các tích phân sau:
a) \(\int_0^2 {\left| {1 - x} \right|} dx\) b) \(\int_0^{{\pi \over 2}} s i{n^2}xdx\)
c) \(\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx\) d) \(\int_0^\pi s in2xco{s^2}xdx\)
Giải
a) Ta có \(1 - x = 0 ⇔ x = 1\).
\(\int_0^2 {\left| {1 - x} \right|} dx = \int_0^1 {\left| {1 - x} \right|} dx + \int_1^2 {\left| {1 - x} \right|} dx\)
\(= - \int_0^1 {(1 - x)} d(1 - x) + \int_1^2 {(x - 1)} d(x - 1)\)
\( = - {{{{(1 - x)}^2}} \over 2}|_0^1 + {{{{(x - 1)}^2}} \over 2}|_1^2 = {1 \over 2} + {1 \over 2} = 1\)
b) \(\int_0^{{\pi \over 2}} s i{n^2}xdx\)
\( = {1 \over 2}\int_0^{{\pi \over 2}} {(1 - cos2x)} dx\)
\( = {1 \over 2}\left( {x - {1 \over 2}sin2x} \right)|_0^{{\pi \over 2}} = {\pi \over 4}\)
c) \(\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx = \int_0^{ln2} {({e^{x + 1}} + {e^{ - x}})} dx\)
\( = ({e^{x + 1}} - {e^{ - x}})|_0^{ln2} = e + {1 \over 2}\)
d) Ta có : \(sin2xcos^2x\) = \({1 \over 2}sin2x(1 + cos2x) = {1 \over 2}sin2x + {1 \over 4}sin4x\)
Do đó : \(\eqalign{
& \int_0^\pi s in2xco{s^2}xdx = \int_0^\pi {({1 \over 2}sin2x + {1 \over 4}sin4x)} dx \cr
& = ( - {1 \over 4}cos2x - {1 \over {16}}cos4x)|_0^\pi \cr
& = - {1 \over 4} - {1 \over {16}} + {1 \over 4} + {1 \over {16}} = 0 \cr} \).
Bài 3 trang 113 -SGK Giải tích 12
Sử dụng phương pháp biến đổi số, tính tích phân:
a) \(\int_{0}^{3}\frac{x^{2}}{(1+x)^{\frac{3}{2}}}dx\) (Đặt \(u= x+1\))
b) \(\int_{0}^{1}\sqrt{1-x^{2}}dx\) (Đặt \(x = sint\) )
c) \(\int_{0}^{1}\frac{e^{x}(1+x)}{1+x.e^{x}}dx\) (Đặt \(u = 1 + x.{e^x}\))
d)\(\int_{0}^{\frac{a}{2}}\frac{1}{\sqrt{a^{2}-x^{2}}}dx\) (Đặt \(x= asint\))
Giải
a) Đặt \(u= x+1 \Rightarrow du = dx\) và \(x = u - 1\).
Khi \(x =0\) thì \(u = 1, x = 3\) thì \(u = 4\). Khi đó :
\(\int_{0}^{3}\frac{x^{2}}{(1+x)^{\frac{3}{2}}}dx\) = \(\int_{1}^{4}\frac{(u-1)^{2}}{u^{\frac{3}{2}}}du =\int_{1}^{4}\frac{u^{2}-2u+1}{u^{\frac{3}{2}}}du\)
= \((\frac{2}{3}u^{\frac{3}{2}}-4.u^{\frac{1}{2}}-2u^{\frac{-1}{2}})|_{1}^{4}=\frac{5}{3}\)
b) Đặt \(x = sint\), \(0<t<\frac{\pi}{2}\). Ta có: \(dx = costdt\)
và \(\sqrt{1-x^{2}}=\sqrt{1-sin^{2}t}= \sqrt{cos^{2}t}=\left | cost \right |= cos t.\)
Khi \(x = 0\) thì \(t = 0\), khi \(x = 1\) thì \(t= \frac{\pi}{2}\) . Khi đó:
\(\int_{0}^{1}\sqrt{1-x^{2}}dx = \int_{0}^{\frac{\pi}{2}}cos^{2}tdt= \frac{1}{2}\int_{0}^{\frac{\pi}{2}}(1+cos2t)dt\)
\(=\frac{1}{2}(t+\frac{1}{2}sin 2t)|_{0}^{\frac{\pi}{2}}=\frac{1}{2}(\frac{\pi}{2}-0)= \frac{\pi}{4}\)
c) Đặt: \(t = 1 + x{e^x} \Rightarrow dt = {e^x}(1 + x)dx\)
Khi \(x = 0 \Rightarrow t = 1\)
Khi \(x = 1 \Rightarrow t = 1 + e\)
Do đó ta có:
\(\int\limits_0^1 {{{{e^x}(1 + x)} \over {1 + x{e^x}}}dx = \int\limits_1^{1 + e} {{{dt} \over t} = {\rm{[}}\ln |t|{\rm{]}}} } \left| {_1^{1 + e} = \ln (1 + e)} \right.\).
d) Đặt \(x = a\sin t \Rightarrow dx = a\cos tdt\)
Đổi cận:
\(\eqalign{
& x = 0 \Rightarrow t = 0 \cr
& x = {a \over 2} \Rightarrow t = {\pi \over 6} \cr} \)
Do đó ta có:
\(\int\limits_0^{{a \over 2}} {{1 \over {\sqrt {{a^2} - {x^2}} }}dx = \int\limits_0^{{\pi \over 6}} {{{a\cos tdt} \over {\sqrt {{a^2} - {a^2}{{\sin }^2}t} }} = \int\limits_0^{{\pi \over 6}} {dt = t\left| {_0^{{\pi \over 6}} = {\pi \over 6}} \right.} } } \).