Find \(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\). (2 marks)
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Find \(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\). (2 marks)
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\(5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\)
| \(\displaystyle \int \dfrac{5}{\sqrt{7-x^2-6 x}} \, dx\) | \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x^2+6x+9)}} \, dx\) | |
| \(=\displaystyle \int \dfrac{5}{\sqrt{16-(x+3)^2}} \, dx\) | ||
| \(=5\,\sin^{-1} \left( \dfrac{x+3}{4} \right) +c\) |
Find \({\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\). (3 marks) --- 8 WORK AREA LINES (style=lined) --- \(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \) \( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)
\(I\)
\(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)
\(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)
\(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)
\(\text{Let}\ \ u=5-4x-x^2 \)
\(I\)
\(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)
\(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)
\(= \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)
Which expression is equal to `int frac{1}{x^2 + 4x + 10}\ dx`?
`A`
| `int frac{1}{x^3 + 4x + 10}\ dx` | `= int frac{1}{(x + 2)^2 + (sqrt6)^2}\ dx` |
| `= frac{1}{6} tan^-1 (frac{x + 2}{sqrt6}) + c` |
`=> \ A`
By completing the square and using the table of standard integrals, find
`int(dx)/(4x^2-4x+10)` (2 marks)
`1/6 tan^-1((2x-1)/(3))+C`
| `int(dx)/(4x^2-4x+10)` | `=int(dx)/(3^2+(2x-1)^2)` |
| `=1/2 int 2/(3^2+(2x-1)^2) \ dx` | |
| `=1/6 tan^-1((2x-1)/(3))+C` |
By completing the square, find `int (dx)/(sqrt (5+4x-x^2))` . (2 marks)
`sin^-1((x-2)/(3))+C`
| `int(dx)/(sqrt(5+4x-x^2))` | `=int(dx)/(sqrt(9-(x^2-4x+4)))` |
| `=int(dx)/(sqrt(3^2-(x-2)^2)` | |
| `=sin^-1((x-2)/(3))+C` |
Find `int (dx)/(x^2 + 10x + 29)` (2 marks)
`1/2 tan^(-1) ((x + 5)/2) + C`
| `int (dx)/(x^2 + 10x + 29)` | `= int (dx)/((x + 5)^2 + 2^2)` |
| `= 1/2 tan^(-1) ((x + 5)/2) + C` |
Find `int(x^2 + 2x)/(x^2 + 2x + 5)\ dx`. (3 marks)
`x – 5/2 tan^(−1) ((x + 1)/2) + c`
| `int(x^2 + 2x)/(x^2 + 2x + 5)\ dx` | `= int((x^2 + 2x + 5) – 5)/(x^2 + 2x + 5)\ dx` |
| `= int 1 – 5/(x^2 + 2x + 5)\ dx` | |
| `= int 1 – 5/(2^2 + (x + 1)^2)\ dx` | |
| `= x – 5/2 tan^(−1) ((x + 1)/2) + c` |
By completing the square, find
`int (dx)/(x^2 - 6x + 13)` (2 marks)
`1/2 tan^-1 ((x – 3)/2) + c`
| `int (dx)/(x^2 – 6x + 13)` | `=int (dx)/((x – 3)^2 + 4)` |
| `=1/2 tan^-1 ((x – 3)/2) + c` |
Evaluate `int_-1^1 1/(5 - 2t + t^2) \ dt.` (3 marks)
`pi/8`
| `int_-1^1 1/{(5 – 2t + t^2)}dt` | `= int_-1^1 1/{(4 + 1 – 2t + t^2)dt}` |
| `= int_-1^1 1/(4 + (t – 1)^2)dt` | |
| `= 1/2[tan^-1 ((t – 1)/2)]_-1^1` | |
| `= 1/2 [tan^-1 0 – tan^(-1)(-1)]` | |
| `= 1/2 [0 – (-pi/4)]` | |
| `= pi/8` |