Bài 12 trang 147 SGK Giải tích 12

Đề bài

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 \(u = \sqrt {1 + \tan x} \) )

Hướng dẫn giải

a) Đặt \(u = \cos ({\pi  \over 3} - 4x)\)

b) Đặt \(x = {3 \over 5}\tan t\)

c) Đặt u = cos x

d) Đặt \(u = \sqrt {1 + \tan x} \)

Lời giải chi tiết

a) Ta có: \(I=\int\limits_0^{\frac{\pi }{{24}}} {\tan \left( {\frac{\pi }{3} - 4x} \right)dx}  = \int\limits_0^{\frac{\pi }{{24}}} {\frac{{\sin \left( {\frac{\pi }{3} - 4x} \right)}}{{\cos \left( {\frac{\pi }{3} - 4x} \right)}}dx} \)

Đặt \(u = \cos \left( {\frac{\pi }{3} - 4x} \right) \Leftrightarrow du = 4\sin \left( {\frac{\pi }{3} - 4x} \right)dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow u = \frac{1}{2}\\x = \frac{\pi }{{24}} \Rightarrow u =\frac{{\sqrt 3 }}{2}\end{array} \right.\)

Khi đó: \(I = \int\limits_{\frac{1}{2}}^{\frac{{\sqrt 3 }}{2}} {\frac{{du}}{{4u}}}  = \left. {\frac{1}{4}\ln \left| u \right|} \right|_{\frac{1}{2}}^{\frac{{\sqrt 3 }}{2}} = \frac{1}{4}\left( {\ln \frac{{\sqrt 3 }}{2} - \ln \frac{1}{2}} \right) = \frac{1}{4}\ln \sqrt 3 \)

b) Đặt \(x = \frac{3}{5}\tan t \Leftrightarrow dx = \frac{3}{{5{{\cos }^2}t}}dt = \frac{3}{5}\left( {{{\tan }^2}t + 1} \right)dt\).

Đổi cận: \(\left\{ \begin{array}{l}x = \frac{{\sqrt 3 }}{5} \Rightarrow t = \frac{\pi }{6}\\x = \frac{3}{5} \Rightarrow t = \frac{\pi }{4}\end{array} \right.\)

\(\begin{array}{l}I = \int\limits_{\frac{{\sqrt 3 }}{5}}^{\frac{3}{5}} {\frac{{dx}}{{9 + 25{x^2}}}} = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{3\left( {{{\tan }^2}t + 1} \right)dt}}{{5\left( {9 + 25.\frac{9}{{25}}{{\tan }^2}t} \right)}}} \\I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{3\left( {{{\tan }^2}t + 1} \right)}}{{5.9\left( {{{\tan }^2}t + 1} \right)}}dt} = \frac{1}{{15}}\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {dt} = \left. {\frac{t}{{15}}} \right|_{\frac{\pi }{6}}^{\frac{\pi }{4}} = \frac{\pi }{{180}}\end{array}\)

 c) Ta có: \(I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}x{{\cos }^4}xdx}  = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} \)

Đặt \(u = \cos x \Rightarrow du =  - \sin xdx\)

Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Leftrightarrow u = 1\\x = \frac{\pi }{2} \Rightarrow u = 0\end{array} \right.\)

\(\begin{array}{l}\Rightarrow I = - \int\limits_1^0 {\left( {1 - {u^2}} \right){u^4}du} = \int\limits_0^1 {\left( {{u^4} - {u^6}} \right)du} \\\,\,\,\,\,\,I = \left. {\left( {\frac{{{u^5}}}{5} - \frac{{{u^7}}}{7}} \right)} \right|_0^1 = \frac{2}{{35}}\end{array}\)

 d) Đặt \(u = \sqrt {1 + \tan x}  \Leftrightarrow {u^2} = 1 + \tan x \Leftrightarrow 2udu = \frac{1}{{{{\cos }^2}x}}dx\).

Đổi cận: \(\left\{ \begin{array}{l}x = - \frac{\pi }{4} \Rightarrow u = 0\\x = \frac{\pi }{4} \Rightarrow u = \sqrt 2 \end{array} \right.\)

\( \Rightarrow I = \int\limits_0^{\sqrt 2 } {u.2udu}  = 2\int\limits_0^{\sqrt 2 } {{u^2}du}  = 2\left. {\frac{{{u^3}}}{3}} \right|_0^{\sqrt 2 } = \frac{2}{3}.2\sqrt 2  = \frac{{4\sqrt 2 }}{3}\)