1,\(\int sin^4xdx\)
2,\(\int cos^23xdx\)
3,\(\int sin^3xcos^2xdx\)
4,\(\int sin^2xcos^5xdx\)
5,\(\int sin^35xcos5xdx\)
6,\(\int sin^23xcos^23xdx\)
7,\(\int_0^{\frac{\pi}{4}}sec^4xdx\)
8,\(\int cot^3xdx\)
9,\(\int\frac{\left(4x+3\right)dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}\)
10,\(\int\frac{dx}{\left(x+2\right)\sqrt{x^2+4x+3}}\)
11,\(\int\frac{9x^2-24x+6}{x^3-5x^2+6x}dx\)
12,\(\int\frac{6x^2-9x+9}{x^3-3x^2}dx\)
giup e voi a
\(I_1=\int\left(\frac{1-cos2x}{2}\right)^2dx=\frac{1}{4}\int\left(1-2cos2x+cos^22x\right)dx\)
\(=\frac{1}{4}\int\left(1-2cos2x+\frac{1}{2}+\frac{1}{2}cos4x\right)dx\)
\(=\frac{1}{4}\int\left(\frac{3}{2}-2cos2x+\frac{1}{2}cos4x\right)dx\)
\(=\frac{1}{4}\left(\frac{3}{2}x-sin2x+\frac{1}{8}sin4x\right)+C\)
\(I_2=\int\left(\frac{1+cos6x}{2}\right)dx=\frac{1}{2}x+\frac{1}{12}sin6x+C\)
\(I_3=\int sin^2x.cos^2x.sinxdx=-\int\left(1-cos^2x\right)cos^2x.d\left(cosx\right)\)
\(=\int\left(cos^4x-cos^2x\right)d\left(cosx\right)=\frac{1}{5}cos^5x-\frac{1}{3}cos^3x+C\)
\(I_4=\int sin^2x.cos^4x.cosxdx=\int sin^2x\left(1-sin^2x\right)^2d\left(sinx\right)\)
\(=\int\left(sin^6x-2sin^4x+sin^2x\right)d\left(sinx\right)\)
\(=\frac{1}{7}sin^7x-\frac{2}{5}sin^5x+\frac{1}{3}sin^3x+C\)
(với câu kiểu như \(I_3;I_4\) nếu bạn chưa quen tính nguyên hàm trực tiếp theo kiểu đưa hàm vào vi phân thì có thể qua 1 bước phụ đặt \(sinx=t\) khi đó nguyên hàm trở thành \(\int\left(t^6-2t^4+t^2\right)dt=...\) rồi trả biến về x như bình thường)
\(I_5=\frac{1}{5}\int sin^35x.d\left(sin5x\right)=\frac{1}{20}sin^45x+C\)
\(I_6=\frac{1}{4}\int\left(2sin3x.cos3x\right)^2dx=\frac{1}{4}\int sin^26xdx\)
\(=\frac{1}{8}\int\left(1-cos12x\right)dx=\frac{1}{8}\left(x-\frac{1}{12}sin12x\right)+C\)
\(I_7=\int\limits^{\frac{\pi}{4}}_0\frac{1}{cos^4x}dx=\int\limits^{\frac{\pi}{4}}_0\frac{1}{cos^2x}.\frac{dx}{cos^2x}\)
Đặt \(tanx=t\Rightarrow\left\{{}\begin{matrix}\frac{1}{cos^2x}dx=dt\\x=0\Rightarrow t=0\\x=\frac{\pi}{4}\Rightarrow t=1\\\frac{1}{cos^2x}=1+tan^2x=1+t^2\end{matrix}\right.\)
\(\Rightarrow I_7=\int\limits^1_0\left(1+t^2\right)dt=\left(t+\frac{1}{3}t^3\right)|^1_0=\frac{4}{3}\)
\(I_8=\int\frac{cos^3x}{sin^3x}dx=\int\frac{cos^2x}{sin^3x}.cosxdx=\int\frac{1-sin^2x}{sin^3x}.cosxdx\)
Đặt \(sinx=t\Rightarrow cosx.dx=dt\)
\(I_8=\int\frac{1-t^2}{t^3}dt=\int\left(t^{-3}-\frac{1}{t}\right)dt=-\frac{1}{2t^2}-ln\left|t\right|+C\)
\(=-\frac{1}{2sin^2x}-ln\left|sinx\right|+C\)
\(I_9=\int\frac{\left(4x+3\right)dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}=4\int\frac{\left(x-1\right)dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}+7\int\frac{dx}{\left(x^2-2x+2\right)^{\frac{3}{2}}}=4I+7J\)
Đặt \(\sqrt{x^2-2x+2}=t\Rightarrow x^2-2x+2=t^2\)
\(\Rightarrow\left(2x-2\right)dx=2t.dt\Rightarrow\left(x-1\right)dx=t.dt\)
\(\Rightarrow I=\int\frac{tdt}{t^3}=\int\frac{1}{t^2}dt=-\frac{1}{t}+C=-\frac{1}{\sqrt{x^2-2x+2}}+C\)
Xét \(J=\int\frac{dx}{\left[\left(x-1\right)^2+1\right]^{\frac{3}{2}}}\)
Đặt \(x-1=tant\Rightarrow dx=\frac{1}{cos^2t}dt\)
\(J=\int\frac{dt}{cos^2t.\left(tan^2t+1\right)^{\frac{3}{2}}}=\int\frac{dt}{cos^2t.\left(\frac{1}{cos^2t}\right)^{\frac{3}{2}}}=\int\frac{dt}{cos^2t.\frac{1}{cos^3t}}=\int cost.dt\)
\(=sint+C\)
Mặt khác \(\left(x-1\right)^2=tan^2t=\frac{sin^2t}{1-sin^2t}\Rightarrow\frac{1}{\left(x-1\right)^2}=\frac{1}{sin^2t}-1\)
\(\Rightarrow\frac{1}{sin^2t}=\frac{1}{\left(x-1\right)^2}+1=\frac{x^2-2x+2}{\left(x-1\right)^2}\Rightarrow\frac{1}{sint}=\frac{\sqrt{x^2-2x+2}}{x-1}\)
\(\Rightarrow sint=\frac{x-1}{\sqrt{x^2-2x+2}}\)
\(\Rightarrow J=\frac{x-1}{\sqrt{x^2-2x+2}}+C\)
\(\Rightarrow I_9=-\frac{4}{\sqrt{x^2-2x+2}}+\frac{7\left(x-1\right)}{\sqrt{x^2-2x+2}}+C=\frac{7x-11}{\sqrt{x^2-2x+2}}+C\)
\(I_{10}=\int\frac{dx}{\left(x+2\right)\sqrt{x^2+4x+3}}=\int\frac{\left(x+2\right)}{\left(x+2\right)^2\sqrt{x^2+4x+3}}dx\)
\(=\int\frac{\left(x+2\right)dx}{\left(x^2+4x+3+1\right)\sqrt{x^2+4x+3}}\)
Đặt \(\sqrt{x^2+4x+3}=t\Rightarrow x^2+4x+3=t^2\)
\(\Rightarrow\left(2x+4\right)dx=2t.dt\Rightarrow\left(x+2\right)dx=t.dt\)
\(I_{10}=\int\frac{dt}{\left(t^2+1\right)t}=\int\left(\frac{1}{t}-\frac{t}{t^2+1}\right)dt=ln\left|t\right|-\frac{1}{2}ln\left(t^2+1\right)+C\)
\(=ln\left(\sqrt{x^2+4x+3}\right)-ln\left|x+2\right|+C\)
\(I_{11}=\int\frac{9x^2-24x+6}{x^3-5x^2+6x}dx=3\int\frac{3x^2-10x+6}{x^3-5x^2+6x}dx+\int\frac{6x-12}{x^3-5x^2+6x}dx\)
\(=3\int\frac{d\left(x^3-5x^2+6x\right)}{x^3-5x^2+6x}+6\int\frac{1}{x\left(x-3\right)}dx\)
\(=3\int\frac{d\left(x^3-5x^2+6x\right)}{x^3-5x^2+6x}+2\int\left(\frac{1}{x-3}-\frac{1}{x}\right)dx\)
\(=3ln\left|x^3-5x^2+6x\right|+2ln\left|\frac{x-3}{x}\right|+C\)
\(I_{12}=\int\frac{6x^2-9x+9}{x^3-3x^2}dx=2\int\frac{3x^2-6x}{x^3-3x^2}dx+3\int\frac{x-3}{x^2\left(x-3\right)}dx+18\int\frac{dx}{x^2\left(x-3\right)}\)
\(=2\int\frac{d\left(x^3-3x^2\right)}{x^3-3x^2}+3\int\frac{dx}{x^2}+2\int\left(\frac{1}{x}-\frac{3}{x^2}-\frac{1}{x-3}\right)dx\)
\(=2ln\left|x^3-3x^2\right|-\frac{9}{x}+2ln\left|\frac{x}{x-3}\right|+C\)
