#1
$\int\frac{tan^{-1}x}{1+x^{2}}\;dx$
$u=tan^{-1}x$
$du=\frac{dx}{1+x^2}$
$\int u\;du=\frac{u^2}{2}+C$
$\frac{(tan^{-1}x)^{2}{}}{2}+C$
#2
$\int sinx*cosx\;dx$
$u= sinx$
$du= cosxdx$
$\int u\;du= \frac{1}{2} u^2+c$
$\frac{1}{2} sin^2x+c$
#3
$\int sin^2 3x\cos 3x\;dx$
$u= 3x$
$du= 3dx$
$\frac{1}{3}du= dx$
$\frac{1}{3}\int sin^2 u\cos u\;du$
$u= sin u$
$du= cos u\;du$
$\frac{1}{3}\int u^2\;du = \frac{1}{3}\;(\frac{1}{3}u^3)+c$
$\frac{1}{9}u^3+c$
$\frac{1}{9}sin^3\;3x+c$
#4
$\int e^x(1+e^x)^{10}\;dx$
$u= 1+e^x$
$du= e^x\;dx$
$\int u^{10}\;du$
$=\frac{u^{11}}{11}$
$=\frac{(1+e^x)^{11}}{11}+C$
#5
$\int \frac{(lnx)^2}{x}\;dx$
$u= lnx$
$du= \frac{1}{x}\;dx$
$\int u^2\;du$
$=\frac{(u)^3}{3}$
$=\frac{(lnx)^3}{3}+C$
#6
$$\int \frac{1}{2x-1}\;dx$$
$$u= 2x-1$$
$$du= 2dx$$
$$\frac{1}{2}du=dx$$
$$\frac{1}{2}\int \frac{1}{u}\;du$$
$$=\frac{1}{2}lnu+C$$
$$=\frac{1}{2}ln(2x-1)+C$$
I don't like maths =P
ResponderEliminarMy god the equations! :)
ResponderEliminarWhat program do you use to write those equations down?
ResponderEliminarI used a script called mathjax. If you whant to use it tell and I show you how
ResponderEliminarWhy? Why would anyone make a blog talking about integrals? I don't suppose you'll be doing Taylor series and 3d integrals now, will ya?
ResponderEliminari will try to solve!
ResponderEliminarLet's see what he will be posting in the future :P
ResponderEliminarI'm so bad in math xD
ResponderEliminar