Question
Evaluate the following integrals:
$\int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$
$\int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$
Therefore,
$7-(\text{x}^2+6\text{x}+9-9)$
$=16-(\text{x}^2+6\text{x}+9)$ $=16-(\text{x}+3)^2$ $=(4)^2-(\text{x}+3)^2$ $\therefore\ \int\frac{1}{\sqrt{7-6\text{x}-\text{x}^2}}\text{ dx}$ $=\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}$ Let x + 3 = t $\Rightarrow\text{dx}=\text{dt}$ $\Rightarrow\int\frac{1}{\sqrt{(4)^2-(\text{x}+3)^2}}\text{ dx}=\int\frac{1}{\sqrt{(4)^2-(\text{t})^2}}\text{ dt}$ $=\sin^{-1}\Big(\frac{\text{t}}{4}\Big)+\text{C}$ $=\sin^{-1}\Big(\frac{\text{x}+3}{4}\Big)+\text{C}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(\text{x}+\text{y})\ \text{dy}+(\text{x}-\text{y})\ \text{dx}=0;\ \text{y}=1\ \text{when}\ \text{x}=1$
$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
Verify that (adjoint A)A = |A|I = A (adjoint A) for the above matrices.