Recent content by Pedro1

Well, I thougt that the dimension was the number of vectors that the trivial basis have. In this case $\mathbb{R^2}$ have $2$.

Thanks for the help. I kown the the vector $u$ is on the trivial basis of $\mathbb{R^2}$ and so $(-9,6)=-9(1,0)+6(0,1)$. Normaly what I do is write each vector of the trivial basis as a linear combination of the given basis, in this case $B$. But, $(1,0)=\lambda\left (-\frac{3}{2},1 \right )$...

Let S be a subspace of $\mathbb{R^2}$, such that $S=\{(x,y):2x+3y=0 \}$. Find a basis,$B$, for $S$ and write $u=(-9,6)$ in the $B$ basis. So, I started to solve $2x+3y=0$ for $x$ and I got $x=-\frac{3}{2}y$. Then I could write, $\left[ \begin{matrix} x \\ y \end{matrix}\right] = \left[...

Hi, this is my first question in this forum. Find $\int \cos(\ln x) \mathrm dx$. I started by substitution. Let $u=\ln x$, so we get $\mathrm du=\frac{1}{x} \mathrm dx$ and $x=e^{u}$. Then the integral stays like this: $$\int \cos(u)e^{u} \mathrm du$$ But now I can't managed how to finish...