a) \(\displaystyle \int\limits_0^1 {{{(y - 1)}^2}\sqrt y } dy\)
Đặt \(\displaystyle t = \sqrt y \Rightarrow {t^2} = y \Rightarrow 2tdt = dy\)
\(\displaystyle \Rightarrow \int\limits_0^1 {{{(y - 1)}^2}\sqrt y } dy\) \(\displaystyle = \int\limits_0^1 {{{\left( {{t^2} - 1} \right)}^2}.t.2tdt} \) \(\displaystyle = 2\int\limits_0^1 {{t^2}\left( {{t^4} - 2{t^2} + 1} \right)dt} \) \(\displaystyle = 2\int\limits_0^1 {\left( {{t^6} - 2{t^4} + {t^2}} \right)dt} \) \(\displaystyle = 2\left. {\left( {\frac{{{t^7}}}{7} - 2.\frac{{{t^5}}}{5} + \frac{{{t^3}}}{3}} \right)} \right|_0^1\) \(\displaystyle = 2\left( {\frac{1}{7} - \frac{2}{5} + \frac{1}{3}} \right) = \frac{{16}}{{105}}\)
b) \(\displaystyle \int\limits_1^2 {({z^2} + 1)\sqrt[3]{{{{(z - 1)}^2}}}} dz\)
Đặt \(\displaystyle u = \sqrt[3]{{{{(z - 1)}^2}}}\) \(\displaystyle \Rightarrow {u^3} = {\left( {z - 1} \right)^2}\) \(\displaystyle \Rightarrow z = 1 + {u^{\frac{3}{2}}} \Rightarrow dz = \frac{3}{2}{u^{\frac{1}{2}}}du\)
\(\displaystyle \Rightarrow \int\limits_1^2 {({z^2} + 1)\sqrt[3]{{{{(z - 1)}^2}}}} dz\) \(\displaystyle = \int\limits_0^1 {\left[ {{{\left( {1 + {u^{\frac{3}{2}}}} \right)}^2} + 1} \right].u.\frac{3}{2}{u^{\frac{1}{2}}}du} \) \(\displaystyle = \frac{3}{2}\int\limits_0^1 {{u^{\frac{3}{2}}}\left( {2 + 2{u^{\frac{3}{2}}} + {u^3}} \right)du} \)
\(\displaystyle = \frac{3}{2}\int\limits_0^1 {\left( {2{u^{\frac{3}{2}}} + 2{u^3} + {u^{\frac{9}{2}}}} \right)du} \) \(\displaystyle = \frac{3}{2}\left. {\left( {2.\frac{2}{5}{u^{\frac{5}{2}}} + 2.\frac{{{u^4}}}{4} + \frac{2}{{11}}{u^{\frac{{11}}{2}}}} \right)} \right|_0^1\) \(\displaystyle = \frac{3}{2}\left( {\frac{4}{5} + \frac{1}{2} + \frac{2}{{11}}} \right) = \frac{{489}}{{220}}\)
c) \(\displaystyle \int\limits_1^e {\frac{{\sqrt {4 + 5\ln x} }}{x}} dx\)
Đặt \(\displaystyle t = \sqrt {4 + 5\ln x} \Rightarrow {t^2} = 4 + 5\ln x\) \(\displaystyle \Rightarrow 2tdt = \frac{5}{x}dx \Rightarrow \frac{{dx}}{x} = \frac{2}{5}tdt\)
\(\displaystyle \Rightarrow \int\limits_1^e {\frac{{\sqrt {4 + 5\ln x} }}{x}} dx\) \(\displaystyle = \int\limits_2^3 {t.\frac{2}{5}tdt} = \frac{2}{5}\int\limits_2^3 {{t^2}dt} \) \(\displaystyle = \frac{2}{5}.\left. {\frac{{{t^3}}}{3}} \right|_2^3 = \frac{2}{5}\left( {\frac{{27}}{3} - \frac{8}{3}} \right) = \frac{{38}}{{15}}\).
d) \(\displaystyle \int\limits_0^{\frac{\pi }{2}} {({{\cos }^5}\varphi } - {\sin ^5}\varphi )d\varphi \)
Phương pháp:
Sử dụng lý thuyết: Nếu \(\displaystyle f\left( x \right)\) xác định và liên tục trên đoạn \(\displaystyle \left[ {a;b} \right]\) thì \(\displaystyle \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin x} \right)dx} = \int\limits_0^{\frac{\pi }{2}} {f\left( {\cos x} \right)dx} \) (bài 3.22 trang 172 SBT Giải tích 12 cơ bản).
Cách giải:
Xét hàm số \(\displaystyle f\left( t \right) = {t^5}\) xác định và liên tục trên \(\displaystyle \mathbb{R}\).
Khi đó \(\displaystyle \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \varphi } \right)d\varphi } = \int\limits_0^{\frac{\pi }{2}} {f\left( {\cos \varphi } \right)d\varphi } \) hay \(\displaystyle \int\limits_0^{\frac{\pi }{2}} {{{\sin }^5}\varphi d\varphi } = \int\limits_0^{\frac{\pi }{2}} {{{\cos }^5}\varphi d\varphi } \)
\(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{2}} {{{\cos }^5}\varphi d\varphi } - \int\limits_0^{\frac{\pi }{2}} {{{\sin }^5}\varphi d\varphi } = 0\) \(\displaystyle \Rightarrow \int\limits_0^{\frac{\pi }{2}} {\left( {{{\cos }^5}\varphi - {{\sin }^5}\varphi } \right)d\varphi } = 0\)
