1. ʃ(cos(\(\frac{\pi}{2}x\))\(-\)\(\frac{2}{6x+5}\))dx
2. ʃ\(2x^3\)\(\sqrt{4-x^4}\)dx
3.ʃ2x\(\sqrt{\varepsilon^{4+x^2}}\)dx
4.ʃx\(\sqrt[3]{1-x^2}\)dx
5.ʃcosx\(\varepsilon^{\sin\chi}\)dx
6.ʃ\(\frac{\cos\chi}{1+sINx}\)dx
7.ʃ(x+1)\(\sqrt{x-1}\)dx
8.ʃ(2 x+1)\(^{ }\)20dx
9.ʃ\(\frac{9x^2}{\sqrt{1-x^3}}dX\)
10.ʃ\(\frac{\chi}{\sqrt{2x+3}}\)dx
\(I_1=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=\frac{2}{\pi}\int cos\left(\frac{\pi x}{2}\right)d\left(\frac{\pi x}{2}\right)-\frac{1}{3}\int\frac{d\left(6x+5\right)}{6x+5}\)
\(=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
\(I_2=-\frac{1}{2}\int\left(4-x^4\right)^{\frac{1}{2}}d\left(4-x^4\right)=-\frac{1}{2}.\frac{\left(4-x^4\right)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{-\sqrt{\left(4-x^4\right)^3}}{3}+C\)
\(I_3=2\int e^{\frac{1}{2}\left(4+x^2\right)}d\left(\frac{1}{2}\left(4+x^2\right)\right)=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^2}}+C\)
\(I_4=-\frac{1}{2}\int\left(1-x^2\right)^{\frac{1}{3}}d\left(1-x^2\right)=-\frac{1}{2}.\frac{\left(1-x^2\right)^{\frac{4}{3}}}{\frac{4}{3}}+C=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
\(I_5=\int e^{sinx}d\left(sinx\right)=e^{sinx}+C\)
\(I_6=\int\frac{d\left(1+sinx\right)}{1+sinx}=ln\left(1+sinx\right)+C\)
\(I_7=\int\left(x+1\right)\sqrt{x-1}dx\)
Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow dx=2tdt\)
\(\Rightarrow I_7=\int\left(t^2+2\right).t.2t.dt=\int\left(2t^4+4t^2\right)dt=\frac{2}{5}t^5+\frac{4}{3}t^3+C\)
\(=\frac{2}{5}\sqrt{\left(1-x\right)^5}+\frac{4}{3}\sqrt{\left(1-x\right)^3}+C\)
\(I_8=\int\left(2x+1\right)^{20}dx\)
Đặt \(2x+1=t\Rightarrow2dx=dt\Rightarrow dx=\frac{1}{2}dt\)
\(\Rightarrow I_8=\frac{1}{2}\int t^{20}dt=\frac{1}{42}t^{21}+C=\frac{1}{42}\left(2x+1\right)^{21}+C\)
\(I_9=-3\int\left(1-x^3\right)^{-\frac{1}{2}}d\left(1-x^3\right)=-3.\frac{\left(1-x^3\right)^{\frac{1}{2}}}{\frac{1}{2}}+C=-6\sqrt{1-x^3}+C\)
\(I_{10}=\int\frac{x}{\sqrt{2x+3}}dx\)
Đặt \(\sqrt{2x+3}=t\Rightarrow x=\frac{1}{2}t^2-\frac{3}{2}\Rightarrow dx=t.dt\)
\(\Rightarrow I_{10}=\int\frac{\frac{1}{2}t^2-\frac{3}{2}}{t}.t.dt=\frac{1}{2}\int\left(t^2-3\right)dt=\frac{2}{3}t^3-\frac{3}{2}t+C\)
\(=\frac{2}{3}\sqrt{\left(2x+3\right)^3}-\frac{3}{2}\sqrt{2x+3}+C\)
a/ \(I=\int cos\left(\frac{\pi x}{2}\right)dx-\int\frac{2}{6x+5}dx=I_1+I_2\)
Xét \(I_1=\int cos\left(\frac{\pi x}{2}\right)dx\)
Đặt \(u=\frac{\pi x}{2}\Rightarrow x=\frac{2}{\pi}u\Rightarrow dx=\frac{2}{\pi}du\)
\(\Rightarrow I_1=\int cosu.\frac{2}{\pi}du=\frac{2}{\pi}\int cosu.du=\frac{2}{\pi}sinu+C=\frac{2}{\pi}sin\left(\frac{\pi x}{2}\right)+C\)
Xét \(I_2=\int\frac{2}{6x+5}dx\)
Đặt \(u=6x+5\Rightarrow x=\frac{1}{6}u-\frac{5}{6}\Rightarrow dx=\frac{1}{6}du\)
\(\Rightarrow I_2=\int\frac{2}{u}.\frac{1}{6}du=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}ln\left|u\right|+C=\frac{1}{3}ln\left|6x+5\right|+C\)
\(\Rightarrow I=I_1+I_2=\frac{2}{\pi}sin\left(\frac{\pi}{2}x\right)-\frac{1}{3}ln\left|6x+5\right|+C\)
b/ Đặt \(\sqrt{4-x^4}=u\Rightarrow x^4=4-u^2\Rightarrow4x^3dx=-2udu\)
\(\Rightarrow2x^3dx=-udu\)
\(\Rightarrow I=\int u.\left(-u.du\right)=-\int u^2du=-\frac{1}{3}u^3+C\)
\(=-\frac{1}{3}\sqrt{\left(4-x^4\right)^3}+C\)
c/ \(I=\int2e^{\frac{1}{2}\left(4+x^2\right)}.xdx\)
Đặt \(u=\frac{1}{2}\left(4+x^2\right)=2+\frac{x^2}{2}\Rightarrow xdx=du\)
\(\Rightarrow I=\int2e^udu=2e^u+C=2e^{\frac{1}{2}\left(4+x^2\right)}+C=2\sqrt{e^{4+x^4}}+C\)
d/ Đặt \(\sqrt[3]{1-x^2}=u\Rightarrow x^2=1-u^3\Rightarrow2xdx=-3u^2du\)
\(\Rightarrow xdx=-\frac{3}{2}u^2du\)
\(\Rightarrow I=\int u.\left(-\frac{3}{2}u^2du\right)=-\frac{3}{2}\int u^3du=-\frac{3}{8}u^4+C\)
\(=-\frac{3}{8}\sqrt[3]{\left(1-x^2\right)^4}+C\)
e/ Đặt \(sinx=u\Rightarrow cosxdx=du\)
\(\Rightarrow I=\int e^udu=e^u+C=e^{sinx}+C\)
f/ Đặt \(1+sinx=u\Rightarrow cosxdx=du\)
\(\Rightarrow I=\int\frac{du}{u}=ln\left|u\right|+C=ln\left|1+sinx\right|+C\)
Do \(1+sinx>0\) khi mẫu xác định nên có trị tuyệt đối hay ko cũng được
Câu 7;8;10 thay t bằng u là được
Câu 9:
Đặt \(\sqrt{1-x^3}=u\Rightarrow x^3=1-u^2\Rightarrow3x^2dx=-2udu\)
\(\Rightarrow9x^2dx=-6udu\)
\(\Rightarrow I=\int\frac{-6udu}{u}=-6\int du=-6u+C=-6\sqrt{1-x^3}+C\)
