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Bài 12. Tính các tích phân sau bằng phương pháp đổi biến số
a) \(\int\limits_0^{{\pi \over 24}} {\tan ({\pi \over 4} – 4x)dx} \) (đặt \(u = \cos ({\pi \over 3} – 4x)\) )
b) \(\int\limits_{{{\sqrt 3 } \over 5}}^{{3 \over 5}} {{{dx} \over {9 + 25{x^2}}}} \) (đặt \(x = {3 \over 5}\tan t\) )
c) \(\int\limits_0^{{\pi \over 2}} {{{\sin }^3}} x{\cos ^4}xdx\) (đặt u = cos x)
d) \(\int\limits_{{{ – \pi } \over 4}}^{{\pi \over 4}} {{{\sqrt {1 + \tan x} } \over {{{\cos }^2}x}}} dx\) (đặt \(u = \sqrt {1 + \tan x} \) )
a) Ta có:
Đặt \(u = \cos ({\pi \over 3} – 4x)\) thì \(u’ = 4sin({\pi \over 3} – 4x)\)
Khi \(x = 0\) thì \(u = {1 \over 2}\) ; khi \(x = {\pi \over {24}} \Rightarrow u = {{\sqrt 3 } \over 2}\)
Khi đó:
\(\eqalign{
& \int\limits_0^{{\pi \over {24}}} {\tan ({\pi \over 3}} – 4x)dx = {1 \over 4}\int\limits_0^{{\pi \over {24}}} {{{d\cos ({\pi \over 3} – 4x)} \over {\cos ({\pi \over 3} – 4x)}}} \cr
& = {1 \over 4}\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{du} \over u}} ={1 \over 4}\ln |u|\left| {_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}} \right.= {1 \over 4}\ln \sqrt 3 \cr} \)
b)
Đặt
\(x = {3 \over 5}\tan t \Rightarrow \left\{ \matrix{
9 + 25{x^2} = 9(1 + {\tan ^2}t) \hfill \cr
dx = {3 \over 5}(1 + {\tan ^2}t) \hfill \cr} \right.\)
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Đổi cận: \(x = {{\sqrt 3 } \over 5} \Rightarrow t = {\pi \over 6};x = {3 \over 5} \Rightarrow t = {\pi \over 4}\)
Do đó:
\(\int\limits_{{{\sqrt 3 } \over 5}}^{{3 \over 5}} {{{dx} \over {9 + 25{x^2}}}} = \int\limits_{{\pi \over 6}}^{{\pi \over 4}} {{1 \over {15}}dt ={1 \over {15}}t\left| {_{{\pi \over 6}}^{{\pi \over 4}}} \right. {\pi \over {180}}} \)
c) Đặt \(t = cos x\) thì \(dt = -sin x dx\)
Khi \(x = 0 \Rightarrow t = 1;x = {\pi \over 2} \Rightarrow t = 0\)
Do đó:
\(\eqalign{
& \int\limits_0^{{\pi \over 2}} {{{\sin }^3}x{{\cos }^4}xdx = \int\limits_1^0 { – (1 – {t^2}){t^4}} dt} \cr
& = – \int\limits_0^1 {({t^4} – {t^6})dt = – ({{{t^5}} \over 5}} – {{{t^7}} \over 7})\left| {_0^1} \right. = {2 \over {35}} \cr} \)
d) Đặt \(u = \sqrt {1 + \tan x} \Rightarrow {t^2} = 1 + \tan x \Rightarrow 2tdt = {{dx} \over {{{\cos }^2}x}}\)
Do đó:
\(\int\limits_{{{ – \pi } \over 4}}^{{\pi \over 4}} {{{\sqrt {1 + \tan x} } \over {{{\cos }^2}x}}} dx = \int\limits_0^{\sqrt 2 } {2{t^2}dt = {2 \over 3}} {t^3}\left| {_0^{\sqrt 2 }} \right. = {{4\sqrt 2 } \over 3}\)