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
- \[\int_{\frac{-1}{2}}{\frac{1}{2}}\sqrt[3]{ [1-x]{2}}dx\] = \[-\int_{\frac{-1}{2}}{\frac{1}{2}}[1-x]{\frac{2}{3}}d[1-x]=-\frac{3}{5}[1-x]{\frac{5}{3}}|_{\frac{-1}{2}}{\frac{1}{2}}\]
\= \[-\frac{3}{5}\left [ \frac{1}{2\sqrt[3]{4}}-\frac{3\sqrt[3]{9}}{2\sqrt[3]{4}} \right ]=\frac{3}{10\sqrt[3]{4}}[3\sqrt[3]{9}-1]\]
- \[\int_{0}{\frac{\pi}{2}}sin[\frac{\pi}{4}-x]dx\]=\[-\int_{0}{\frac{\pi}{2}}sin[\frac{\pi}{4}-x]d[\frac{\pi}{4}-x]\] = \[cos[\frac{\pi}{4}-x]|_{0}^{\frac{\pi}{2}}\]
\= \[cos[\frac{\pi}{4}-\frac{\pi}{2}]-cos\frac{\pi}{4}=0\]
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:
- \[\int_0^2 {\left| {1 - x} \right|} dx\] b] \[\int_0^{{\pi \over 2}} s i{n^2}xdx\]
- \[\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx\] d] \[\int_0^\pi s in2xco{s^2}xdx\]
Giải
- Ta có \[1 - x = 0 ⇔ x = 1\].
\[\int_0^2 {\left| {1 - x} \right|} dx = \int_0^1 {\left| {1 - x} \right|} dx + \int_1^2 {\left| {1 - x} \right|} dx\]
\[= - \int_0^1 {[1 - x]} d[1 - x] + \int_1^2 {[x - 1]} d[x - 1]\]
\[ = - {{{{[1 - x]}^2}} \over 2}|_0^1 + {{{{[x - 1]}^2}} \over 2}|_1^2 = {1 \over 2} + {1 \over 2} = 1\]
- \[\int_0^{{\pi \over 2}} s i{n^2}xdx\]
\[ = {1 \over 2}\int_0^{{\pi \over 2}} {[1 - cos2x]} dx\]
\[ = {1 \over 2}\left[ {x - {1 \over 2}sin2x} \right]|_0^{{\pi \over 2}} = {\pi \over 4}\]
- \[\int_0^{ln2} {{{{e^{2x + 1}} + 1} \over {{e^x}}}} dx = \int_0^{ln2} {[{e^{x + 1}} + {e^{ - x}}]} dx\]
\[ = [{e^{x + 1}} - {e^{ - x}}]|_0^{ln2} = e + {1 \over 2}\]
- 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:
- \[\int_{0}{3}\frac{x{2}}{[1+x]^{\frac{3}{2}}}dx\] [Đặt \[u= x+1\]]
- \[\int_{0}{1}\sqrt{1-x{2}}dx\] [Đặt \[x = sint\] ]
- \[\int_{0}{1}\frac{e{x}[1+x]}{1+x.e^{x}}dx\] [Đặt \[u = 1 + x.{e^x}\]]
d]\[\int_{0}{\frac{a}{2}}\frac{1}{\sqrt{a{2}-x^{2}}}dx\] [Đặt \[x= asint\]]
Giải
- Đặt \[u= x+1 \Rightarrow du = dx\] và \[x = u - 1\].
Khi \[x =0\] thì \[u = 1, x = 3\] thì \[u = 4\]. Khi đó :
\[\int_{0}{3}\frac{x{2}}{[1+x]{\frac{3}{2}}}dx\] = \[\int_{1}{4}\frac{[u-1]{2}}{u{\frac{3}{2}}}du =\int_{1}{4}\frac{u{2}-2u+1}{u^{\frac{3}{2}}}du\]
\= \[[\frac{2}{3}u^{\frac{3}{2}}-4.u^{\frac{1}{2}}-2u^{\frac{-1}{2}}]|_{1}^{4}=\frac{5}{3}\]
- Đặt \[x = sint\], \[0