Mọi người ơi , giúp e tính tích phân bất định với ạ ! Cảm ơn m.n ạ !
a.\(\int\frac{x+6}{\sqrt{x^2-2x+10}}dx\)
b.\(\int\frac{x}{\sqrt{3-2x-x^2}}dx\)
c.\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)
d,\(\int\frac{dx}{1+tanx}\)
e.\(\int tan^3xdx\)
f. \(\int cos^3xdx\)
g. \(\int sin^2x.cos^3xdx\)
h. \(\int sinx.cos2xdx\)
i. \(\int\frac{sin2x}{1+cos^2x}dx\)
a.
\(I=\int\frac{\frac{1}{2}\left(2x-2\right)+7}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx+7\int\frac{1}{\sqrt{x^2-2x+10}}dx=\frac{1}{2}I_1+7I_2\)
Xét \(I_1=\int\frac{2x-2}{\sqrt{x^2-2x+10}}dx=\int\frac{d\left(x^2-2x+10\right)}{\sqrt{x^2-2x+10}}=2\sqrt{x^2-2x+10}+C_1\)
Xét \(I_2=\int\frac{dx}{\sqrt{x^2-2x+10}}=\int\frac{dx}{\sqrt{\left(x-1\right)^2+9}}\)
Đặt
\(u=x-1+\sqrt{\left(x-1\right)^2+10}\Rightarrow du=\left(1+\frac{\left(x-1\right)}{\sqrt{\left(x-1\right)^2+10}}\right)dx=\frac{x-1+\sqrt{\left(x-1\right)^2+10}}{\sqrt{\left(x-1\right)^2+10}}dx\)
\(\Rightarrow du=\frac{u}{\sqrt{\left(x-1\right)^2+10}}dx\Rightarrow\frac{dx}{\sqrt{\left(x-1\right)^2+10}}=\frac{du}{u}\)
\(\Rightarrow I_2=\int\frac{du}{u}=ln\left|u\right|+C_2=ln\left|x-1+\sqrt{x^2-2x+10}\right|+C_2\)
\(\Rightarrow I=\sqrt{x^2-2x+10}+7ln\left|x-1+\sqrt{x^2-2x+10}\right|+C\)
2.
\(I=\int\frac{\frac{1}{2}\left(2x+2\right)-1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx-\int\frac{1}{\sqrt{3-2x-x^2}}dx=\frac{1}{2}I_1-I_2\)
Xét \(I_1=\int\frac{2x+2}{\sqrt{3-2x-x^2}}dx=-\int\frac{d\left(3-2x-x^2\right)}{\sqrt{3-2x-x^2}}=-2\sqrt{3-2x-x^2}+C_1\)
Xét \(I_2=\int\frac{1}{\sqrt{3-2x-x^2}}dx=\int\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\)
Đặt \(x+1=2sinu\Rightarrow dx=2cosu.du\)
\(\Rightarrow I_2=\int\frac{2cosu.du}{2.cosu}=\int du=u+C_2=arcsin\left(\frac{x+1}{2}\right)+C_2\)
\(\Rightarrow I=-\sqrt{3-2x-x^2}-arcsin\left(\frac{x+1}{2}\right)+C\)
c/
\(I=\int\frac{1-\sqrt{x}}{\sqrt{1-x}}dx\)
Đặt \(\sqrt{x}=sint\Rightarrow x=sin^2t\Rightarrow dx=2sint.cost.dt\)
\(\Rightarrow I=\int\frac{2sint.cost\left(1-sint\right)}{\sqrt{1-sin^2t}}dt=\int\frac{2sint.cost\left(1-sint\right)}{cost}dt=\int\left(2sint-2sin^2t\right)dt\)
\(=\int\left(2sint+cos2t-1\right)dt=-2cost+\frac{1}{2}sin2t-t+C\)
\(=-2\sqrt{1-sin^2t}+\frac{1}{2}sint\sqrt{1-sin^2t}-t+C\)
\(=-2\sqrt{1-x}+\frac{1}{2}\sqrt{x\left(1-x\right)}-arcsin\left(\sqrt{x}\right)+C\)
4.
\(I=\int\frac{cosx}{sinx+cosx}dx=\frac{1}{2}\int\frac{cosx-sinx+cosx+sinx}{sinx+cosx}dx\)
\(=\frac{1}{2}\int\frac{cosx-sinx}{sinx+cosx}dx+\int dx=\frac{1}{2}\int\frac{d\left(sinx+cosx\right)}{sinx+cosx}+\int dx\)
\(=ln\left|sinx+cosx\right|+x+C\)
5.
\(I=\int tan^3xdx=\int\frac{sin^3x}{cos^3x.dx}=\int\frac{\left(1-cos^2x\right).sinx}{cos^3x}dx\)
Đặt \(cosx=t\Rightarrow sinx.dx=-dt\)
\(I=\int\frac{t^2-1}{t^3}dt=\int\left(\frac{1}{t}-\frac{1}{t^3}\right)dt=ln\left|t\right|+\frac{1}{2t^2}+C\)
\(=ln\left|cosx\right|+\frac{1}{2cos^2x}+C\)
6.
\(I=\int cos^3xdx=\int\left(1-sin^2x\right)cosxdx\)
\(=\int\left(1-sin^2x\right)d\left(sinx\right)=sinx-\frac{1}{3}sin^3x+C\)
7.
\(I=\int sin^2x.cos^3xdx=\int sin^2x\left(1-sin^2x\right)cosxdx\)
\(=\int\left(sin^2x-sin^4x\right)d\left(sinx\right)=\frac{1}{3}sin^3x-\frac{1}{5}sin^5x+C\)
8.
\(I=\int sinx.cos2xdx=\int\left(2cos^2x-1\right)sinxdx\)
\(=\int\left(1-2cos^2x\right)d\left(cosx\right)=cosx-\frac{2}{3}cos^3x+C\)
9.
\(I=\int\frac{sin2x}{1+cos^2x}dx=-\int\frac{2\left(-sinx\right).cosx}{1+cos^2x}dx=-\int\frac{d\left(cos^2x\right)}{1+cos^2x}\)
\(=-ln\left|1+cos^2x\right|+C\)