Tổng hợp lý thuyết bài tập tìm nguyên hàm bằng cách đổi biến số (đặt x = hàm theo biến t) toán lớp 12

Bài tập tìm nguyên hàm bằng cách đổi biến số (Đặt x = hàm theo biến t)

Phương pháp đổi biến số đặt x = hàm theo biến t

@ Mẫu 1: Nếu $f\left( x \right)$ có chứa $\sqrt{{{a}^{2}}-{{x}^{2}}}$ ta đặt $x=a\sin t\,\,\left( t\in \left[ -\frac{\pi }{2};\frac{\pi }{2} \right] \right)$

 $\Rightarrow \left\{ \begin{array}  {} dx=a\cos tdt \\  {} \sqrt{{{a}^{2}}-{{a}^{2}}{{\sin }^{2}}t}=\left| a \right|\cos t \\ \end{array} \right.$

@ Mẫu 2: Dạng $\sqrt{{{x}^{2}}+{{a}^{2}}}$ thì đổi biến số $x=a\tan t,\,\,\,\,\left( t\in \left( -\frac{\pi }{2};\frac{\pi }{2} \right) \right)$

 $\Rightarrow \left\{ \begin{array}  {} dx=\frac{adt}{{{\cos }^{2}}t} \\  {} \sqrt{{{a}^{2}}+{{x}^{2}}}=\sqrt{{{a}^{2}}+{{a}^{2}}{{\tan }^{2}}t}=\frac{\left| a \right|}{\cos t} \\ \end{array} \right.$

@ Mẫu 3: Dạng $\sqrt{{{x}^{2}}-{{a}^{2}}}$ thì ta đặt $x=\frac{a}{\sin t}$  (hoặc $x=\frac{a}{\cos t}$).

$\Rightarrow \left\{ \begin{array}  {} dx=\frac{-a\cos tdt}{{{\sin }^{2}}t} \\  {} \sqrt{{{x}^{2}}-{{a}^{2}}}=\sqrt{{{a}^{2}}{{\cot }^{2}}t} \\ \end{array} \right.$

@ Mẫu 4: Dạng $\int{\frac{dx}{{{x}^{2}}+{{a}^{2}}}}$ thì ta đặt $x=a\tan t.$

@ Mẫu 5: Nếu $f\left( x \right)$ có chứa $\sqrt{\frac{a+x}{a-x}}$ thì đặt $x=a\cos 2t\xrightarrow{\,}\left\{ \begin{array}  {} dx=d\left( a\cos 2t \right)=-2a.\sin 2tdt \\  {} \sqrt{\frac{a+x}{a-x}}=\sqrt{\frac{1+\cos 2t}{1-\cos 2t}}=\sqrt{\frac{{{\cos }^{2}}t}{{{\sin }^{2}}t}} \\ \end{array} \right.$

Ü Một số kết quả quan trọng cần lưu ý khi giải trắc nghiệm:

Ø $\int{\frac{dx}{{{x}^{2}}+{{a}^{2}}}}=\frac{1}{a}\arctan \frac{x}{a}+C\,\,\,\,\left( a\ne 0 \right)$

Ø $\int{\frac{dx}{{{x}^{2}}-{{a}^{2}}}}=\frac{1}{2a}\ln \left| \frac{x+a}{x-a} \right|+C$

Ø $\int{\frac{dx}{\sqrt{{{x}^{2}}+a}}}=\ln \left| x+\sqrt{{{x}^{2}}+a} \right|+C\,\,\,\,\left( a\ne 0 \right)$

Ø $\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}=\arcsin \frac{x}{a}+C\,\,\,\,\,\left( a>0 \right)$

Bài tập trắc nghiệm tìm nguyên hàm bằng cách đổi biến số có đáp án chi tiết

Lời giải chi tiết

a) Đặt $x=2\sin t\Rightarrow \left\{ \begin{array}  {} dx=d\left( 2\sin t \right)=2\cos tdt \\  {} \sqrt{4-{{x}^{2}}}=\sqrt{4-4{{\sin }^{2}}t}=2\cos t \\ \end{array} \right.\xrightarrow{\,}{{I}_{1}}=\int{\frac{dx}{\sqrt{4-{{x}^{2}}}}}=\int{\frac{2\cos tdt}{2\cos t}}=\int{dt}=t+C$

Từ phép đặt $x=2\sin t\Leftrightarrow t=\arcsin \left( \frac{x}{2} \right)\xrightarrow{\,\,}{{I}_{1}}=\arcsin \left( \frac{x}{2} \right)+C$

b) Đặt $x=\sin t\Rightarrow \left\{ \begin{array}  {} dx=d\left( \sin t \right)=\cos tdt \\  {} \sqrt{1-{{x}^{2}}}=\sqrt{1-{{\sin }^{2}}t}=\cos t \\ \end{array} \right.$

Khi đó ${{I}_{2}}=\int{\sqrt{1-{{x}^{2}}}dx}=\int{\cos t.\cos tdt}=\int{\frac{1+\cos 2t}{2}dt}=\frac{1}{2}\int{dt}+\frac{1}{2}\int{\cos 2tdt}=\frac{t}{2}+\frac{1}{4}\sin 2t+C.$

Từ $x=\sin t\Rightarrow \left\{ \begin{array}  {} \cos t=\sqrt{1-{{\sin }^{2}}t}=\sqrt{1-{{x}^{2}}} \\  {} t=\arcsin x \\ \end{array} \right.\xrightarrow{\,}\sin 2t=2\sin t.\cos t=2x\sqrt{1-{{x}^{2}}}$

$\xrightarrow{\,}{{I}_{2}}=\frac{\arcsin x}{2}+\frac{1}{2}x\sqrt{1-{{x}^{2}}}+C$

c) Đặt $x=\sin t\xrightarrow{\,}\left\{ \begin{array}  {} dx=d\left( \sin t \right)=\cos tdt \\  {} \sqrt{1-{{x}^{2}}}=\sqrt{1-{{\sin }^{2}}t}=\cos t \\ \end{array} \right.$

Khi đó, ${{I}_{3}}=\int{\frac{{{x}^{2}}dx}{\sqrt{1-{{x}^{2}}}}}=\int{\frac{{{\sin }^{2}}t.\cos tdt}{\cos t}}=\int{{{\sin }^{2}}tdt}=\int{\frac{1-\cos 2t}{2}dt}=\frac{1}{2}t-\frac{1}{4}\sin 2t+C$

Từ $x=\sin t\Rightarrow \left\{ \begin{array}  {} \cos t=\sqrt{1-{{\sin }^{2}}t}=\sqrt{1-{{x}^{2}}} \\  {} t=\arcsin x \\ \end{array} \right.\xrightarrow{{}}\sin 2t=2\sin t.\cos t=2x\sqrt{1-{{x}^{2}}}$

$\xrightarrow{\,}{{I}_{3}}=\frac{\arcsin x}{2}-\frac{1}{2}x\sqrt{1-{{x}^{2}}}+C$

d) Đặt $x=3\sin t\xrightarrow{\,}\left\{ \begin{array}  {} dx=d\left( 3\sin t \right)=3\cos tdt \\  {} \sqrt{9-{{x}^{2}}}=\sqrt{9-9{{\sin }^{2}}t}=3\cos t \\ \end{array} \right.$

${{I}_{4}}=\int{{{x}^{2}}\sqrt{9-{{x}^{2}}}dx}=\int{9{{\sin }^{2}}t.3\cos tdt}=81\int{{{\sin }^{2}}t.{{\cos }^{2}}tdt}=\frac{81}{4}\int{{{\sin }^{2}}2tdt}=\frac{81}{4}\int{\frac{1-\cos 4t}{2}dt}$

 $=\frac{81}{4}\left[ \frac{1}{2}\int{dt}-\frac{1}{2}\int{\cos 4tdt} \right]=\frac{81}{4}\left( \frac{t}{2}-\frac{1}{8}\sin 4t \right)+C$

Từ $x=3\sin t\Rightarrow \left\{ \begin{array}  {} \cos t=\sqrt{1-{{\sin }^{2}}t}=\sqrt{1-\frac{{{x}^{2}}}{9}} \\  {} t=\arcsin \left( \frac{x}{3} \right) \\ \end{array} \right.\xrightarrow{\,}\sin 2t=\frac{2x}{3}\sqrt{1-\frac{{{x}^{2}}}{9}}$

Mà $\cos 2t=1-2{{\sin }^{2}}t=1-2{{\left( \frac{x}{3} \right)}^{2}}=1-\frac{2{{x}^{2}}}{9}\xrightarrow{\,}\sin 4t=2\sin 2t.\cos 2t=2.\frac{2x}{3}\sqrt{1-\frac{{{x}^{2}}}{9}}.\left( 1-\frac{2{{x}^{2}}}{9} \right)$

Từ đó ta được ${{I}_{4}}=\frac{81}{4}\left[ \frac{\arcsin \left( \frac{x}{3} \right)}{2}-\frac{x}{6}\sqrt{1-\frac{{{x}^{2}}}{9}}.\left( 1-\frac{2{{x}^{2}}}{9} \right) \right]+C.$

Lời giải chi tiết

a) Đặt $x=\tan x\xrightarrow{\,\,}\left\{ \begin{array}  {} dx=d\left( \tan t \right)=\frac{dt}{{{\cos }^{2}}t}=\left( 1+{{\tan }^{2}}t \right)dt \\  {} 1+{{x}^{2}}=1+{{\tan }^{2}}t \\ \end{array} \right.\xrightarrow{\,}{{I}_{1}}=\int{\frac{\left( 1+{{\tan }^{2}}t \right)dt}{1+{{\tan }^{2}}t}}=\int{dt}=t+C$

Từ giả thiết đặt $x=\tan t\Leftrightarrow t=\arctan x\xrightarrow{\,}{{I}_{1}}=\arctan x+C.$

b) Ta có ${{I}_{2}}=\int{\sqrt{{{x}^{2}}+2x+5}dx}=\int{\sqrt{{{\left( x+1 \right)}^{2}}+4}d\left( x+1 \right)}\xrightarrow{t=x+1}I=\int{\sqrt{{{t}^{2}}+4}dt}$

Đặt $t=2\tan u\xrightarrow{\,\,}\left\{ \begin{array}  {} dt=d\left( 2\tan u \right)=\frac{2du}{{{\cos }^{2}}u} \\  {} \sqrt{4+{{t}^{2}}}=\sqrt{4+4{{\tan }^{2}}u}=\frac{2}{\cos u} \\ \end{array} \right.\xrightarrow{\,\,}{{I}_{2}}=\int{\frac{2du}{\frac{2}{\cos u}.{{\cos }^{2}}u}}=\int{\frac{du}{\cos u}}=\int{\frac{\cos udu}{{{\cos }^{2}}u}}$

$=\int{\frac{d\left( \sin u \right)}{1-{{\sin }^{2}}u}}=\frac{1}{2}\int{\frac{\left( 1+\sin u \right)+\left( 1-\sin u \right)}{\left( 1+\sin u \right)\left( 1-\sin u \right)}d\left( \sin u \right)}=\frac{1}{2}\int{\frac{d\left( \sin u \right)}{1-\sin u}}+\frac{1}{2}\int{\frac{d\left( \sin u \right)}{1+\sin u}}=\frac{1}{2}\ln \left| \frac{1+\sin u}{1-\sin u} \right|+C.$

Từ phép đặt $t=2\tan u\Leftrightarrow \tan u=\frac{t}{2}\xrightarrow{\,}\frac{1}{{{\cos }^{2}}u}=1+\frac{{{t}^{2}}}{4}\xrightarrow{\,}{{\sin }^{2}}u=1-{{\cos }^{2}}u=1-\frac{4}{4+{{t}^{2}}}=\frac{{{t}^{2}}}{4+{{t}^{2}}}$

Từ đó ta được ${{I}_{2}}=\frac{1}{2}\ln \left| \frac{1+\sin u}{1-\sin u} \right|+C=\frac{1}{2}\ln \left| \frac{1+\frac{t}{\sqrt{4+{{t}^{2}}}}}{1-\frac{t}{\sqrt{4+{{t}^{2}}}}} \right|+C=\frac{1}{2}\ln \left| \frac{1+\frac{x+1}{\sqrt{{{x}^{2}}+2x+5}}}{1-\frac{x+1}{\sqrt{{{x}^{2}}+2x+5}}} \right|+C.$

c) Đặt $x=2\tan t\xrightarrow{\,}\left\{ \begin{array}  {} dx=d\left( 2\tan t \right)=\frac{2dt}{{{\cos }^{2}}t}=2\left( 1+{{\tan }^{2}}t \right)dt \\  {} \sqrt{{{x}^{2}}+4}=\sqrt{4{{\tan }^{2}}t+4} \\ \end{array} \right.$

$\Rightarrow {{I}_{3}}=\int{\frac{4{{\tan }^{2}}t.2\left( 1+{{\tan }^{2}}t \right)dt}{2\sqrt{1+{{\tan }^{2}}t}}}=4\int{{{\tan }^{2}}t\sqrt{1+{{\tan }^{2}}t}dt}=4\int{\frac{{{\sin }^{2}}t}{{{\cos }^{3}}t}dt}$

 $=4\int{\frac{{{\sin }^{2}}t.\cos tdt}{{{\cos }^{4}}t}}=4\int{\frac{{{\sin }^{2}}t.d\left( \sin t \right)}{{{\left( 1-{{\sin }^{2}}t \right)}^{2}}}}$

Đặt $u=\sin t\xrightarrow{\,}{{I}_{3}}=4\int{\frac{{{u}^{2}}}{{{\left( 1-{{u}^{2}} \right)}^{2}}}du}=4\int{{{\left( \frac{u}{1-u} \right)}^{2}}du}=4\int{{{\left[ \frac{1}{2}\frac{\left( 1+u \right)-\left( 1-u \right)}{\left( 1+u \right)\left( 1-u \right)} \right]}^{2}}du}$

$=\int{{{\left( \frac{1}{1-u}-\frac{1}{1+u} \right)}^{2}}du}=\int{\frac{du}{{{\left( 1-u \right)}^{2}}}}+\int{\frac{du}{{{\left( 1+u \right)}^{2}}}}-\int{\frac{2du}{\left( 1-u \right)\left( 1+u \right)}}$

$=-\int{\frac{d\left( 1-u \right)}{{{\left( 1-u \right)}^{2}}}}+\int{\frac{d\left( 1+u \right)}{{{\left( 1+u \right)}^{2}}}}-\int{\frac{\left( 1-u \right)\left( 1+u \right)du}{\left( 1-u \right)\left( 1+u \right)}}$

$=-\frac{1}{1-u}-\frac{1}{1+u}-\int{\left( \frac{1}{1+u}+\frac{1}{1-u} \right)du}=-\frac{1}{1-u}-\frac{1}{1+u}-\int{\frac{du}{1+u}}-\int{\frac{du}{1-u}}$

$=-\frac{1}{1-u}-\frac{1}{1+u}-\ln \left| 1+u \right|+\ln \left| u-1 \right|+C=\frac{1}{u-1}-\frac{1}{1+u}+\ln \left| \frac{u-1}{u+1} \right|+C$

$\xrightarrow{\,}{{I}_{3}}=\frac{1}{u-1}-\frac{1}{1+u}+\ln \left| \frac{u-1}{u+1} \right|+C=\frac{1}{\sin t-t}-\frac{1}{\sin t+1}+\ln \left| \frac{\sin t-1}{\sin t+1} \right|+C.$

Lại có $x=2\tan t\Leftrightarrow \tan t=\frac{x}{2}\xrightarrow{\,}\frac{1}{{{\cos }^{2}}t}=1+{{\tan }^{2}}t=1+\frac{{{x}^{2}}}{4}\Leftrightarrow {{\cos }^{2}}t=\frac{4}{4+{{x}^{2}}}\xrightarrow{\,}{{\sin }^{2}}t=\frac{{{x}^{2}}}{4+{{x}^{2}}}$

$\Leftrightarrow \sin t=\frac{x}{\sqrt{4+{{x}^{2}}}}\xrightarrow{\,}{{I}_{3}}=\frac{1}{\frac{x}{\sqrt{4+{{x}^{2}}}}-1}-\frac{1}{\frac{x}{\sqrt{4+{{x}^{2}}}}+1}+\ln \left| \frac{\frac{x}{\sqrt{4+{{x}^{2}}}}-1}{\frac{x}{\sqrt{4+{{x}^{2}}}}+1} \right|+C.$

Lời giải chi tiết

a) Đặt $x=\frac{1}{\sin t}\Rightarrow \left\{ \begin{array}  {} dx=d\left( \frac{1}{\sin t} \right)=\frac{-\cos tdt}{{{\sin }^{2}}t} \\  {} \sqrt{{{x}^{2}}-1}=\sqrt{\frac{1}{{{\sin }^{2}}t}-1} \\ \end{array} \right.\Leftrightarrow \left\{ \begin{array}  {} dx=\frac{-\cos tdt}{{{\sin }^{2}}t} \\  {} \sqrt{{{x}^{2}}-1}=\cot t \\ \end{array} \right.$

$\Rightarrow {{I}_{1}}=\int{\frac{dx}{\sqrt{{{x}^{2}}-1}}}=\int{\frac{-\cos tdt}{{{\sin }^{2}}t.\cot t}}=-\int{\frac{\sin tdt}{{{\sin }^{2}}t}}=\int{\frac{d\left( \cos t \right)}{1-{{\cos }^{2}}t}}$

$=\int{\frac{d\left( \cos t \right)}{\left( 1-\cos t \right)\left( 1+\cos t \right)}}=\frac{1}{2}\int{\frac{\left( 1-\cos t \right)+\left( 1+\cos t \right)}{\left( 1-\cos t \right)+\left( 1+\cos t \right)}d\left( \cos t \right)}=\frac{1}{2}\ln \left| \frac{1+\cos t}{1-\cos t} \right|+C.$

Từ phép đặt $x=\frac{1}{\sin t}\Rightarrow {{\cos }^{2}}t=1-{{\sin }^{2}}t=1-\frac{1}{{{x}^{2}}}\Leftrightarrow \cos t=\frac{\sqrt{{{x}^{2}}-1}}{x}\Rightarrow {{I}_{1}}=\frac{1}{2}\ln \left| \frac{1+\frac{\sqrt{{{x}^{2}}-1}}{x}}{1-\frac{\sqrt{{{x}^{2}}-1}}{x}} \right|+C.$

b) Đặt $x=\frac{2}{\sin t}\xrightarrow{\,}\left\{ \begin{array}  {} dx=d\left( \frac{2}{\sin t} \right)=\frac{-2\cos tdt}{{{\sin }^{2}}t} \\  {} \sqrt{{{x}^{2}}-4}=\sqrt{\frac{4}{{{\sin }^{2}}t}-4} \\ \end{array} \right.\overset{\,}{\longleftrightarrow}\left\{ \begin{array}  {} dx=\frac{-2\cos tdt}{{{\sin }^{2}}t} \\  {} \sqrt{{{x}^{2}}-4}=2\cos t\Rightarrow {{x}^{2}}\sqrt{{{x}^{2}}-4}=\frac{8\cot t}{{{\sin }^{2}}t} \\ \end{array} \right.$

Khi đó, ${{I}_{2}}=\int{\frac{dx}{{{x}^{2}}\sqrt{{{x}^{2}}-4}}=\int{\frac{-2\cos tdt}{{{\sin }^{2}}t.\frac{8\cot t}{{{\sin }^{2}}t}}}=-\frac{1}{4}\int{\sin tdt}=\frac{1}{4}\cos t}+C.$

Từ  $x=\frac{2}{\sin t}\xrightarrow{\,}{{\cos }^{2}}t=1-{{\sin }^{2}}t=1-\frac{4}{{{x}^{2}}}\Leftrightarrow \cos t=\frac{\sqrt{{{x}^{2}}-4}}{x}\xrightarrow{\,}{{I}_{2}}=\frac{\sqrt{{{x}^{2}}-4}}{4x}+C.$

c) ${{I}_{3}}=\int{\frac{dx}{\sqrt{{{x}^{2}}-2x-2}}}=\int{\frac{d\left( x-1 \right)}{\sqrt{{{\left( x-1 \right)}^{2}}-3}}}\xrightarrow{t=x-1}{{I}_{3}}=\int{\frac{dt}{\sqrt{{{t}^{2}}-3}}}=\int{\frac{dt}{\sqrt{{{t}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}}}.$

Đặt $t=\frac{\sqrt{3}}{\sin u}\xrightarrow{\,}\left\{ \begin{array}  {} dt=d\left( \frac{\sqrt{3}}{\sin u} \right)=\frac{-\sqrt{3}\cos udu}{{{\sin }^{2}}u} \\  {} \sqrt{{{t}^{2}}-3}=\sqrt{\frac{3}{{{\sin }^{2}}u}-3} \\ \end{array} \right.\overset{\,}{\longleftrightarrow}\left\{ \begin{array}  {} dt=\frac{-\sqrt{3}\cos udu}{{{\sin }^{2}}u} \\  {} \sqrt{{{t}^{2}}-3}=\sqrt{3}\cot u \\ \end{array} \right.$

$\xrightarrow{\,}{{I}_{3}}=\int{\frac{dt}{\sqrt{{{t}^{2}}-3}}}=\int{\frac{-\sqrt{3}\cos udu}{{{\sin }^{2}}u.\sqrt{3}\cot u}}=-\int{\frac{\sin udu}{{{\sin }^{2}}u}}=\int{\frac{d\left( \cos u \right)}{1-{{\cos }^{2}}u}}=\int{\frac{d\left( \cos u \right)}{\left( 1-\cos u \right)\left( 1+\cos u \right)}}$

$=\frac{1}{2}\int{\frac{\left( 1-\cos u \right)+\left( 1+\cos u \right)}{\left( 1-\cos u \right)\left( 1+\cos u \right)}d\left( \cos u \right)}=\frac{1}{2}\ln \left| \frac{1+\cos u}{1-\cos u} \right|+C.$

$t=\frac{\sqrt{3}}{\sin u}\Rightarrow {{\cos }^{2}}u=1-\frac{3}{{{t}^{2}}}\Leftrightarrow \cos t=\frac{\sqrt{{{t}^{2}}-3}}{t}\Rightarrow {{I}_{3}}=\frac{1}{2}\ln \left| \frac{1+\frac{\sqrt{{{t}^{2}}-3}}{t}}{1-\frac{\sqrt{{{t}^{2}}-3}}{t}} \right|+C=\frac{1}{2}\ln \left| \frac{1+\frac{\sqrt{{{x}^{2}}-2x-2}}{x-1}}{1-\frac{\sqrt{{{x}^{2}}-2x-2}}{x-1}} \right|+C.$

Lời giải chi tiết

Ta có: $x=\sin t\,\,\left( t\in \left[ -\frac{\pi }{2};\frac{\pi }{2} \right] \right)\Rightarrow \left\{ \begin{array}  {} dx=\cos tdt \\  {} \sqrt{1-{{x}^{2}}}=\sqrt{1-{{\sin }^{2}}t}=\left| \cos t \right|=\cos t \\ \end{array} \right.$

Khi đó $I=\int{{{\sin }^{2}}t.{{\cos }^{2}}tdt}=\frac{1}{4}\int{{{\sin }^{2}}2tdt}=\frac{1}{8}\int{\left( 1-\cos 4t \right)dt}\Rightarrow I=\frac{t}{8}-\frac{\sin 4t}{32}+C.$ Chọn C.

Lời giải chi tiết

Ta có $I=\int{\sqrt{\frac{9}{{{\cos }^{2}}t}-9}d\left( \frac{3}{\cos t} \right)}=\int{3\sqrt{\frac{1}{{{\cos }^{2}}t}-1}.\frac{-9}{{{\cos }^{2}}t}.\left( -\sin t \right)dt}$

 $=9\int{\sqrt{\frac{{{\sin }^{2}}t}{{{\cos }^{2}}t}}.\frac{\sin t}{{{\cos }^{2}}t}dt}=9\int{\frac{{{\sin }^{2}}t}{{{\cos }^{3}}}dt}.$ Chọn B.

Lời giải chi tiết

Đặt $x=2\sin t\,\,\left( t\in \left( -\frac{\pi }{2};\frac{\pi }{2} \right) \right)\Rightarrow \left\{ \begin{array}  {} dx=2\cos tdt \\  {} \sqrt{4-{{x}^{2}}}=\sqrt{4-{{\sin }^{2}}t}=2\left| \cos t \right|=2\cos t \\ \end{array} \right.$

Khi đó $I=\int{\frac{2\cos tdt}{2\cos t}}=\int{dt=t+C}=\arcsin \frac{x}{2}+C.$ Chọn A.

Tổng quát: $\int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}=\arcsin \frac{x}{a}+C\,\,\left( a>0 \right)$

Lời giải chi tiết

Ta có: $I=\int{\frac{dx}{\sqrt{1-x-{{x}^{2}}}}}=\int{\frac{dx}{\sqrt{\frac{5}{4}-{{\left( \frac{1}{2}+x \right)}^{2}}}}}=\int{\frac{d\left( x+\frac{1}{2} \right)}{\sqrt{\frac{5}{4}-{{\left( \frac{1}{2}+x \right)}^{2}}}}}$

 $=\arcsin \frac{x+\frac{1}{2}}{\frac{\sqrt{5}}{2}}+C=\arcsin \frac{2x+1}{\sqrt{5}}+C.$ Chọn D.

Lời giải chi tiết

Đặt $x=\sin t\,\,\left( t\in \left( -\frac{\pi }{2};\frac{\pi }{2} \right) \right)\Rightarrow \left\{ \begin{array}  {} dx=\cos tdt \\  {} \sqrt{1-{{x}^{2}}}=\sqrt{1-{{\sin }^{2}}t}=\left| \cos t \right|=\cos t \\ \end{array} \right..$

Khi đó: $I=\int{\frac{{{\sin }^{2}}t.\cos tdt}{{{\cos }^{5}}t}}=\int{\frac{{{\sin }^{2}}t}{{{\cos }^{2}}t}.\frac{1}{{{\cos }^{2}}t}dt}=\int{{{\tan }^{2}}td\left( \tan t \right)}=\frac{{{\tan }^{3}}t}{3}+C.$ Chọn B.

Lời giải chi tiết

Đặt $x=\cos 2t\Rightarrow dx=-2\sin 2tdt=-4\sin t\cos tdt.$

Mặt khác $\sqrt{\frac{1+\cos 2t}{1-\cos 2t}}=\sqrt{\frac{2{{\cos }^{2}}t}{2{{\sin }^{2}}t}}=\sqrt{\frac{{{\cos }^{2}}t}{{{\sin }^{2}}t}}=\left| \frac{\cos t}{\sin t} \right|=\frac{\cos t}{\sin t}$

Khi đó $I=\int{\frac{\cos t}{\sin t}.\left( -4\sin t\cos t \right)dt}=-4\int{{{\cos }^{2}}tdt.}$ Chọn A.

Lời giải chi tiết

Đặt $x=\frac{1}{\cos t}\,\,\left( t\in \left( 0;\frac{\pi }{2} \right) \right)\Rightarrow dx=\frac{-{{\left( \cos t \right)}^{\prime }}}{{{\cos }^{2}}t}dt=\frac{\sin t}{{{\cos }^{2}}t}dt$

Lại có: $\sqrt{\frac{1}{{{\cos }^{2}}t}-1}=\sqrt{{{\tan }^{2}}t}=\left| \tan t \right|=\tan t$

Do đó $I=\int{\frac{\tan t}{\frac{1}{\cos t}}.\frac{\sin t}{{{\cos }^{2}}t}dt}=\int{{{\tan }^{2}}tdt}.$ Chọn B.

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