Nguyên hàm của hàm số lượng giác (Bài 2)

NGUYÊN HÀM CỦA HÀM SỐ LƯỢNG GIÁC (BÀI 2)

A. Các dạng toán và bài tập mẫu (Tiếp)

V. Dạng 5. $I=\int{\frac{dx}{a.{{\sin }^{2}}x+b.\sin x\cos x+c.{{\cos }^{2}}x}}$

1. Phương pháp tính

$I=\int{\frac{dx}{\left( a{{\tan }^{2}}x+b\tan x+c \right).{{\cos }^{2}}x}}$

Đặt $\tan x=t\Rightarrow \frac{dx}{{{\cos }^{2}}x}=dt$. Suy ra $I=\int{\frac{dt}{a{{t}^{2}}+bt+c}}$

2. Ví dụ áp dụng

•$I=\int{\frac{dx}{3{{\sin }^{2}}x-2\sin x\cos x-{{\cos }^{2}}x}}=\int{\frac{dx}{\left( 3{{\tan }^{2}}x-2\tan x-1 \right){{\cos }^{2}}x}}$

Đặt $\tan x=t\Rightarrow \frac{dx}{{{\cos }^{2}}x}=dt$

$\Rightarrow I=\int{\frac{dt}{3{{t}^{2}}-2t-1}}=\int{\frac{dt}{\left( t-1 \right)\left( 3t+1 \right)}}$

                $=\frac{1}{4}\int{\left( \frac{1}{t-1}-\frac{3}{3t+1} \right)dt}=\frac{1}{4}\int{\frac{dt}{t-1}}-\frac{1}{4}\int{\frac{d\left( 3t+1 \right)}{3t+1}}$

                 $=\frac{1}{4}\ln \left| \frac{t-1}{3t+1} \right|+C=\frac{1}{4}\ln \left| \frac{\tan x-1}{3\tan x+1} \right|+C$

•$J=\int{\frac{dx}{{{\sin }^{2}}x-2\sin x\cos x-2{{\cos }^{2}}x}}=\int{\frac{dx}{\left( {{\tan }^{2}}x-2\tan x-2 \right){{\cos }^{2}}x}}$

Đặt $\tan x=t\Rightarrow \frac{dx}{{{\cos }^{2}}x}=dt$

$\Rightarrow J=\int{\frac{dt}{{{t}^{2}}-2t-2}}=\int{\frac{d\left( t-1 \right)}{{{\left( t-1 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}}=\frac{1}{2\sqrt{3}}\ln \left| \frac{t-1-\sqrt{3}}{t-1+\sqrt{3}} \right|+C$

         $=\frac{1}{2\sqrt{3}}\ln \left| \frac{\tan x-1-\sqrt{3}}{\tan x-1+\sqrt{3}} \right|+C$

VI. Dạng 6. $I=\int{\frac{{{a}_{1}}\sin x+{{b}_{1}}\cos x}{{{a}_{2}}\sin x+{{b}_{2}}\cos x}dx}$

1. Phương pháp tính

Ta tìm $A,\,B$ sao cho:

${{a}_{1}}\sin x+{{b}_{1}}\cos x=A\left( {{a}_{2}}\sin x+{{b}_{2}}\cos x \right)+B\left( {{a}_{2}}\cos x-{{b}_{2}}\sin x \right)$

2. Ví dụ áp dụng

•$I=\int{\frac{4\sin x+3\cos x}{\sin x+2\cos x}dx}$

Ta tìm $A,\,B$ sao cho:

$4\sin x+3\cos x=A\left( \sin x+2\cos x \right)+B\left( \cos x-2\sin x \right)$

Từ đó: $I=\int{\frac{2\left( \sin x+2\cos x \right)-\left( \cos x-2\sin x \right)}{\sin x+2\cos x}}dx$

               $=2\int{dx}-\int{\frac{d\left( \sin x+2\cos x \right)}{\sin x+2\cos x}}=2x-\ln \left| \sin x+2\cos x \right|+C$

•$J=\int{\frac{3\cos x-2\sin x}{\cos x-4\sin x}}dx$

Ta tìm $A,\,B$ sao cho:

$3\cos x-2\sin x=A\left( \cos x-4\sin x \right)+B\left( -\sin x-4\cos x \right)$

$\Rightarrow 3\cos x-2\sin x=\left( A-4B \right)\cos x+\left( -4A-B \right)\sin x$

Từ đó: $J=\int{\frac{\frac{11}{17}\left( \cos x-4\sin x \right)-\frac{10}{17}\left( -\sin x-4\cos x \right)}{\cos x-4\sin x}}dx$

$=\frac{11}{17}\int{dx}-\frac{10}{17}\int{\frac{d\left( \cos x-4\sin x \right)}{\cos x-4\sin x}}=\frac{11}{17}x-\frac{10}{17}\ln \left| \cos x-4\sin x \right|+C$

3. Chú ý

1. Nếu gặp $I=\int{\frac{{{a}_{1}}\sin x+{{b}_{1}}\cos x}{{{\left( {{a}_{2}}\sin x+{{b}_{2}}\cos x \right)}^{2}}}}dx$ ta vẫn tìm $A,\,B$ sao cho:

${{a}_{1}}\sin x+{{b}_{1}}\cos x=A\left( {{a}_{2}}\sin x+{{b}_{2}}\cos x \right)+B\left( {{a}_{2}}\cos x-{{b}_{2}}\sin x \right)$

2. Nếu gặp $I=\int{\frac{{{a}_{1}}\sin x+{{b}_{1}}\cos x+{{c}_{1}}}{{{a}_{2}}\sin x+{{b}_{2}}\cos x+{{c}_{2}}}}dx$ ta tìm $A,\,B$ sao cho:

${{a}_{1}}\sin x+{{b}_{1}}\cos x+{{c}_{1}}=A\left( {{a}_{2}}\sin x+{{b}_{2}}\cos x+{{c}_{2}} \right)+B\left( {{a}_{2}}\cos x-{{b}_{2}}\sin x \right)+C$

Chẳng hạn:

•$I=\int{\frac{8\cos x}{{{\left( \sqrt{3}\sin x+\cos x \right)}^{2}}}dx}$

Ta tìm $A,\,B$ sao cho:

$8\cos x=A\left( \sqrt{3}\sin x+\cos x \right)+B\left( \sqrt{3}\cos x-\sin x \right)$

$\Rightarrow 8\cos x=\left( A\sqrt{3}-B \right)\sin x+\left( A+B\sqrt{3} \right)\cos x$

Từ đó: $I=\int{\frac{2\left( \sqrt{3}\sin x+\cos x \right)+2\sqrt{3}\left( \sqrt{3}\cos x-\sin x \right)}{{{\left( \sqrt{3}\sin x+\cos x \right)}^{2}}}}dx$

$=2\int{\frac{dx}{\sqrt{3}\sin x+\cos x}}+2\sqrt{3}\int{\frac{d\left( \sqrt{3}\sin x+\cos x \right)}{{{\left( \sqrt{3}\sin x+\cos x \right)}^{2}}}}=2{{I}_{1}}-\frac{2\sqrt{3}}{\sqrt{3}\sin x+\cos x}+C$

Tìm ${{I}_{1}}=\int{\frac{dx}{\sqrt{3}\sin x+\cos x}}=\frac{1}{2}\int{\frac{dx}{\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x}}=\frac{1}{2}\int{\frac{dx}{\sin x\cos \frac{\pi }{6}+\cos x\sin \frac{\pi }{6}}}$

      $=\frac{1}{2}\int{\frac{dx}{\sin \left( x+\frac{\pi }{6} \right)}}=\frac{1}{2}\int{\frac{d\left( x+\frac{\pi }{6} \right)}{\sin \left( x+\frac{\pi }{6} \right)}}=\frac{1}{2}\ln \left| \tan \frac{x+\frac{\pi }{6}}{2} \right|+C=\frac{1}{2}\ln \left| \tan \left( \frac{x}{2}+\frac{\pi }{12} \right) \right|+C$

Vậy $I=\ln \left| \tan \left( \frac{x}{2}+\frac{\pi }{12} \right) \right|-\frac{2\sqrt{3}}{\sqrt{3}\sin x+\cos x}+C$

• $J=\int{\frac{8\sin x+\cos x+5}{2\sin x-\cos x+1}dx}$

Ta tìm $A,\,B,\,C$ sao cho:

$8\sin x+\cos x+5=A\left( 2\sin x-\cos x+1 \right)+B\left( 2\cos x+\sin x \right)+C$

$\Rightarrow 8\sin x+\cos x+5=\left( 2A+B \right)\sin x+\left( -A+2B \right)\cos x+A+C$

Từ đó: $J=\int{\frac{3\left( 2\sin x-\cos x+1 \right)+2\left( 2\cos x+\sin x \right)+2}{2\sin x-\cos x+1}}dx$

$=3\int{dx}+2\int{\frac{2\cos x+\sin x}{2\sin x-\cos x+1}}dx+2\frac{dx}{2\sin x-\cos x+1}$

$=3x+2\ln \left| 2\sin x-\cos x+1 \right|+2{{J}_{1}}$

Tìm ${{J}_{1}}=\int{\frac{dx}{2\sin x-\cos x+1}}$

$\Rightarrow {{J}_{1}}=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2.\frac{2t}{1+{{t}^{2}}}-\frac{1-{{t}^{2}}}{1+{{t}^{2}}}+1}}=\int{\frac{dt}{{{t}^{2}}+2t}}=\int{\frac{dt}{t\left( t+2 \right)}}=\frac{1}{2}\int{\left( \frac{1}{t}-\frac{1}{t+2} \right)}dt$

          $=\frac{1}{2}\ln \left| \frac{t}{t+2} \right|+C=\frac{1}{2}\ln \left| \frac{\tan \frac{x}{2}}{\tan \frac{x}{2}+2} \right|+C$

Vậy: $J=3x+2\ln \left| 2\sin x-\cos x+1 \right|+\ln \left| \frac{\tan \frac{x}{2}}{\tan \frac{x}{2}+2} \right|+C$

VII. Dạng 7. Biến đổi đưa về nguyên hàm cơ bản hoặc 6 dạng ở trên

• $I=\int{\cos 3x\cos 4xdx}=\frac{1}{2}\int{\left( \cos x+\cos 7x \right)dx}$

      $=\frac{1}{2}\int{\cos xdx}+\frac{1}{2}\int{\cos 7xdx}=\frac{1}{2}\sin x+\frac{1}{14}\sin 7x+C$

• $I=\int{\frac{dx}{\sin x{{\cos }^{3}}x}}=\int{\frac{dx}{\tan x{{\cos }^{4}}x}}=\int{\frac{1}{\tan x}.\frac{1}{{{\cos }^{2}}x}.\frac{dx}{{{\cos }^{2}}x}}=\int{\frac{1}{\tan x}\left( 1+{{\tan }^{2}}x \right)\frac{dx}{{{\cos }^{2}}x}}$

Đặt $\tan x=t\Rightarrow \frac{dx}{{{\cos }^{2}}x}=dt$

• $I=\int{\frac{dx}{{{\sin }^{3}}x}}$

Tính ${{I}_{1}}=\int{\frac{{{\cos }^{2}}x}{{{\sin }^{3}}x}dx}=\int{\frac{1-{{\sin }^{2}}x}{{{\sin }^{3}}x}}dx=\int{\frac{dx}{{{\sin }^{3}}x}-\int{\frac{dx}{\sin x}}}=I-\ln \left| \tan \frac{x}{2} \right|+C$

$\Rightarrow I=-\frac{\cot x}{\sin x}-{{I}_{1}}=-\frac{\cot x}{\sin x}-I+\ln \left| \tan \frac{x}{2} \right|+C$

$\Rightarrow 2I=\ln \left| \tan \frac{x}{2} \right|-\frac{\cot x}{\sin x}+C\Rightarrow I=\frac{1}{2}\ln \left| \tan \frac{x}{2} \right|-\frac{\cot x}{2\sin x}+C$

B. Bài tập tự luyện

1.$I=\int{\frac{dx}{\sin \left( x+\frac{\pi }{3} \right)\cos \left( x+\frac{\pi }{12} \right)}}$   Đáp án:$\sqrt{2}\ln \left| \frac{\sin \left( x+\frac{\pi }{3} \right)}{\cos \left( x+\frac{\pi }{12} \right)} \right|+C$

2.$K=\int{\frac{dx}{\sin x+\tan x}}$                     Đáp án:$\frac{1}{2}\ln \left| \tan \frac{x}{2} \right|-\frac{1}{4}{{\tan }^{2}}\frac{x}{2}+C$

3.$I=\int{\frac{dx}{{{\cos }^{3}}x}}$                                  Đáp án:$\frac{1}{3}\left[ \frac{\tan x}{\cos x}+\ln \left| \tan \left( \frac{x}{2}+\frac{\pi }{4} \right) \right| \right]+C$

4.$I=\int{\frac{\sin 3x\sin 4x}{\tan x+\tan 2x}dx}$                      Đáp án:$-\frac{1}{28}\cos 7x-\frac{1}{20}\cos 5x-\frac{1}{12}\cos 3x-\frac{1}{4}\cos x+C$

5.$I=\int{\tan \left( x+\frac{\pi }{3} \right)\cot \left( x+\frac{\pi }{6} \right)}dx$          Đáp án:$\frac{\sqrt{3}}{3}\ln \left| \frac{\sin \left( x+\frac{\pi }{6} \right)}{\cos \left( x+\frac{\pi }{3} \right)} \right|+x+C$

6.$I=\int{\cos x\sin 2x\cos 3xdx}$            Đáp án:$\frac{1}{2}\int{\sin 2x\left( \cos 2x+\cos 4x \right)}dx$

7.$I=\int{\left( {{\sin }^{3}}x\cos 3x+{{\cos }^{3}}x\sin 3x \right)dx}$Đáp án:$-\frac{3}{4}\int{\sin 2xdx}=\frac{3}{8}\cos 2x+C$

8.\[I=\int{\frac{dx}{{{\sin }^{4}}x\cos x}}\]                          Đáp án:\[-\frac{1}{3{{\sin }^{3}}x}-\frac{1}{\sin x}-\frac{1}{2}\ln \left| \frac{\sin x-1}{\sin x+1} \right|+C\]

9.$I=\int{{{\sin }^{3}}x\sin 3xdx}$                   Đáp án:\[\frac{3}{16}\sin 2x-\frac{3}{32}\sin 4x+\frac{1}{48}\sin 6x-\frac{1}{8}x+C\]

10.$I=\int{\tan x\tan \left( \frac{\pi }{3}-x \right)\tan \left( \frac{\pi }{3}+x \right)dx}$ Đáp án:$-\frac{1}{3}\ln \left| \cos 3x \right|+C$

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