Bài 20: Tính
a) \(\int\limits_0^\pi {5{{\left( {5 – 4\cos t} \right)}^{{1 \over 4}}}} \sin tdt;\)
b) \(\int\limits_0^{\sqrt 3 } {{{{x^3}dx} \over {\sqrt {{x^2} + 1} }}} .\)
a) Đặt \(u = 5 – 4\cos t \Rightarrow du = 4\sin tdt \Rightarrow \sin tdt = {1 \over 4}du\)
\(\int\limits_0^\pi {5{{\left( {5 – 4\cos t} \right)}^{{1 \over 4}}}} \sin tdt = {5 \over 4}\int\limits_1^9 {{u^{{1 \over 4}}}du = \left. {{u^{{5 \over 4}}}} \right|} _1^9 \)
\(= {9^{{5 \over 4}}} – 1\)
b) Đặt \(u = \sqrt {{x^2} + 1} \Rightarrow {u^2} = {x^2} + 1 \)
\(\Rightarrow udu = xdx \Rightarrow {x^3}dx = {x^2}.xdx = \left( {{u^2} – 1} \right)udu\)
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\(\int\limits_0^{\sqrt 3 } {{{{x^3}dx} \over {\sqrt {{x^2} + 1} }}} = \int\limits_1^2 {{{\left( {{u^2} – 1} \right)u} \over u}} du\)
\(\int\limits_1^2 {\left( {{u^2} – 1} \right)du} = \left. {\left( {{{{u^3}} \over 3} – u} \right)} \right|_1^2 = {4 \over 3}\)
Bài 21: Giả sử F là một nguyên hàm của hàm số \(y = {{{\mathop{\rm s}\nolimits} {\rm{inx}}} \over x}\) trên khoảng \(\left( {0; + \infty } \right).\) Khi đó \(\int\limits_1^3 {{{\sin 2x} \over x}} dx\) là
\(\left( A \right)\,\,F\left( 3 \right) – F\left( 1 \right);\) \(\left( B \right)\,F\left( 6 \right) – F\left( 2 \right);\)
\(\left( C \right)\,F\left( 4 \right) – F\left( 2 \right);\) \(\left( D \right)\,F\left( 6 \right) – F\left( 4 \right);\)
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Đặt \(u = 2x \Rightarrow du = 2dx \Rightarrow dx = {1 \over 2}du\)
\(\int\limits_1^3 {{{\sin 2x} \over x}} dx = \int\limits_2^6 {{{\sin u} \over u}} du = \left. {F\left( u \right)} \right|_2^6 = F\left( 6 \right) – F\left( 2 \right).\) chọn (B).
Bài 22: Chứng minh rằng:
a) \(\int\limits_0^1 {f\left( x \right)} dx = \int\limits_0^1 {f\left( {1 – x} \right)dx.} \)
b) \(\int\limits_{ – 1}^1 {f\left( x \right)} dx = \int\limits_0^1 {\left[ {f\left( x \right) + f\left( { – x} \right)} \right]} dx.\)
a) Đặt \(u = 1 – x \Rightarrow du = – dx\)
\(\int\limits_0^1 {f\left( x \right)} dx = \int\limits_1^0 {f\left( {1 – u} \right)} \left( { – du} \right) = \int\limits_0^1 {f\left( {1 – u} \right)} du\)
\(= \int\limits_0^1 {f\left( {1 – x} \right)} dx\)
b) \(\int\limits_{ – 1}^1 {f\left( x \right)} dx = \int\limits_{-1}^0 {f\left( x \right)} dx + \int\limits_0^1 {f\left( x \right)} dx\) với \(\int\limits_{ – 1}^0 {f\left( x \right)} dx\)
Đặt \(u = – x \Rightarrow du = – dx\)
Khi đó \(\int\limits_{ – 1}^0 {f\left( x \right)dx = \int\limits_1^0 {f\left( { – u} \right)} } \left( { – du} \right) = \int\limits_0^1 {f\left( { – u} \right)} du \)
\(= \int\limits_0^1 {f\left( { – x} \right)} dx\)
Do đó \(\int\limits_{ – 1}^1 {f\left( x \right)} dx = \int\limits_0^1 {\left[ {f\left( x \right) + f\left( { – x} \right)} \right]} dx\)
