- LG a
- LG b
Tính
LG a
\[\int\limits_0^\pi {5{{\left[ {5 - 4\cos t} \right]}^{{1 \over 4}}}} \sin tdt;\]
Lời giải chi tiết:
Đặt \[u = 5 - 4\cos t \Rightarrow du = 4\sin tdt \] \[\Rightarrow \sin tdt = {1 \over 4}du\]
\[\int\limits_0^\pi {5{{\left[ {5 - 4\cos t} \right]}^{{1 \over 4}}}} \sin tdt\] \[= \dfrac{5}{4}.\left. {\dfrac{{{u^{\frac{1}{4} + 1}}}}{{\frac{1}{4} + 1}}} \right|_1^9\]\[ = {5 \over 4}\int\limits_1^9 {{u^{{1 \over 4}}}du = \left. {{u^{{5 \over 4}}}} \right|} _1^9 = {9^{{5 \over 4}}} - 1\]
LG b
\[\int\limits_0^{\sqrt 3 } {{{{x^3}dx} \over {\sqrt {{x^2} + 1} }}} .\]
Lời giải chi tiết:
Đặt \[\displaystyle u = \sqrt {{x^2} + 1} \Rightarrow {u^2} = {x^2} + 1 \] \[\displaystyle \Rightarrow udu = xdx \] \[\displaystyle \Rightarrow {x^3}dx = {x^2}.xdx = \left[ {{u^2} - 1} \right]udu\]
\[\displaystyle \int\limits_0^{\sqrt 3 } {{{{x^3}dx} \over {\sqrt {{x^2} + 1} }}} = \int\limits_1^2 {{{\left[ {{u^2} - 1} \right]u} \over u}} du\]
\[\displaystyle \int\limits_1^2 {\left[ {{u^2} - 1} \right]du} = \left. {\left[ {{{{u^3}} \over 3} - u} \right]} \right|_1^2\] \[\displaystyle = \frac{8}{3} - 2 - \frac{1}{3} + 1\]\[\displaystyle = {4 \over 3}\]