Aussie Maths & Science Teachers: Save your time with SmarterEd
Using the substitution \(u=\cos (\theta), \ \dfrac{1}{2} \displaystyle \int_0^{\tfrac{\pi}{2}} \dfrac{\sin (2 \theta)}{1+\cos (\theta)} \, d \theta\) can be expressed as
\(D\)
\(u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta\)
\(\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0\)
\(\text{When}\ \ \theta=0, u=1\)
| \(I\) | \(=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta\) |
| \(=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta\) | |
| \(=\displaystyle \int_1^0-\frac{u}{1+u} \, d u\) | |
| \(=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u\) | |
| \(=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u\) |
\(\Rightarrow D\)
Consider the integral \(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\).
The substitution \(x=5 \sin \theta\) is applied.
Which of the following is obtained?
\(\Rightarrow B\)
\(x=5\, \sin \theta\)
\(\dfrac{dx}{d \theta}=5\, \cos \theta \ \Rightarrow \ dx=5\, \cos \theta \, d \theta\)
\(\text{When} \ \ x=\dfrac{5}{2} \ \Rightarrow \ \sin\, \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)
\(\text{When} \ \ x=-\dfrac{5}{2} \ \Rightarrow \ \sin \theta=-\dfrac{1}{2} \ \Rightarrow \ \theta=-\dfrac{\pi}{6}\)
| \(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\) | \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\left(\dfrac{1}{25-25\, \sin ^2 \theta}\right) \cdot 5\ \cos \theta \, d \theta\) |
| \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{5\, \cos \theta}{25\, \cos ^2 \theta} \, d \theta\) | |
| \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{1}{\cos \theta} \, d \theta\) | |
| \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \, d \theta\) |
\(\Rightarrow B\)
Using the substitution \(x=\cos 2 \theta\), show
\(\displaystyle \int \sqrt{\frac{1-x}{1+x}}\,dx=\sqrt{1-x^2}-\cos ^{-1} x+c\) (4 marks)
--- 10 WORK AREA LINES (style=lined) ---
\(\text{See worked solutions}\)
\(x=\cos 2 \theta\)
\(\dfrac{dx}{d \theta}=-2 \sin 2 \theta \ \Rightarrow\ \ dx=-2 \sin 2 \theta\, d \theta\)
| \(\displaystyle \int \sqrt{\frac{1-x}{1+x}} \, dx\) | \(=\displaystyle\int \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \times -2 \sin 2 \theta\, d \theta\) |
| \(=\displaystyle\int \sqrt{\frac{1-\left(2 \cos ^2 \theta-1\right)}{1+\left(2 \cos ^2 \theta-1\right)}} \times -4 \sin \theta \, \cos \theta \, d \theta\) | |
| \(=\displaystyle \int \sqrt{\frac{2\left(1-\cos ^2 \theta\right)}{2 \cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\) | |
| \(=\displaystyle \int\sqrt{\dfrac{\sin ^2 \theta}{\cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\) | |
| \(=\displaystyle \int-4 \sin ^2 \theta \, d \theta\) | |
| \(=-4 \displaystyle \int \frac{1-\cos 2 \theta}{2} \,d \theta\) | |
| \(=-2 \displaystyle \int 1-\cos 2 \theta \, d \theta\) | |
| \(=-2 \displaystyle \int 1\, d \theta+2 \int \cos 2 \theta \, d \theta\) | |
| \(=-2 \theta+\sin 2 \theta+c\) | |
| \(=-\cos ^{-1} x+\sqrt{1-\cos ^2 2 \theta}+c \quad \text {(note:}\ \ x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x \text{)}\) | |
| \(=-\cos ^{-1} x+\sqrt{1-x^2}+c\) |
Use the substitution \(u=\cos\,x\) to evaluate
\(\displaystyle {\int}_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin x\, dx\). (3 marks)
--- 7 WORK AREA LINES (style=lined) ---
\(1-\dfrac{1}{e}\)
\(u=\cos x\)
\(\dfrac{d u}{d x}=-\sin\,x\ \Rightarrow\ du=-\sin\,x\,dx\)
\(\text {When} \ \ x=\pi, u=-1\)
\(\text{When} \ \ x=\dfrac{\pi}{2}, u=0\)
| \(\displaystyle \int_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin\,x\,d x\) | \(=\displaystyle -\int_0^{-1} e^u\,d u\) |
| \(=\left[-e^u\right]_0^{-1}\) | |
| \(=-e^{-1}+e^0\) | |
| \(=1-\dfrac{1}{e}\) |
Using the substitution \(u=e^x+2 e^{-x}\), and considering \(u^2\), find \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\). (3 marks) --- 8 WORK AREA LINES (style=lined) --- \(\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\) \(u=e^x+2 e^{-x} \ \Rightarrow \ u^2=\left(e^x+2 e^{-x}\right)^2=e^{2 x}+4+4 e^{-2 x}\) \(\dfrac{du}{dx}=e^x-2 e^{-x} \ \Rightarrow \ du=\left(e^x-2 e^{-x}\right)\, d x\)
Show Answers Only
Show Worked Solution
Mean mark 55%.
\(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\)
\(=\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}} \times \frac{e^{-2 x}}{e^{-2 x}}\, d x\)
\(=\displaystyle \int \frac{e^x-2 e^{-x}}{4 e^{-2 x}+8+e^{2 x}}\, d x\)
\(=\displaystyle \int \frac{e^x-2 e^{-x}}{4+\left(e^{2 x}+4+4 e^{-2 x}\right)}\, d x\)
\(=\displaystyle \int \frac{1}{4+u^2}\, d u\)
\(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{u}{2}\right)+c\)
\(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)
--- 1 WORK AREA LINES (style=lined) ---
--- 12 WORK AREA LINES (style=lined) ---
i. `d/(d theta) (sin^3 theta) = 3 cos theta sin^2 theta`
ii. `text(Let)\ x = tan theta`
`(dx)/(d theta) = sec^2 theta \ => \ dx = sec^2 theta\ d theta`
`text(When)\ x = 1, \ theta = pi/4`
`text(When)\ x = 0, \ theta = 0`
| `int_0^1 (x^2)/(1 + x^2)^(5/2) dx` | `= int_0^(pi/4) (tan^2 theta)/((1 + tan^2 theta)^(5/2)) xx sec^2 theta\ d theta` |
| `= int_0^(pi/4) (tan^2 theta)/((sec^2 theta)^(5/2)) xx sec^2 theta\ d theta` | |
| `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/((sec^2 theta)^(3/2))\ d theta` | |
| `= int_0^(pi/4) (sin^2 theta)/(cos^2 theta) · 1/(sec^3 theta)\ d theta` | |
| `= int_0^(pi/4) sin^2 theta cos theta\ d theta` | |
| `= 1/3[sin^3 theta]_0^(pi/4)` | |
| `= 1/3(sin^3\ pi/4 – 0)` | |
| `= 1/3 (1/sqrt2)^3` | |
| `= 1/(6sqrt2)` | |
| `= sqrt2/12` |
Use the substitution `u = cos^2 x` to evaluate `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`ln {:10/9`
| `u` | `= cos^2 x` |
| `(du)/(dx)` | `= -2 sin x cos x` |
| `= -sin 2x` | |
| `du` | `= -sin 2x\ dx` |
`text(When)\ \ x = pi/4,\ \ u = 1/2`
`text(When)\ \ x = 0,\ \ u = 1`
| `:. int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx` | `= -int_1^(1/2) (du)/(4 + u)` |
| `= -[ln (4 + u)]_1^(1/2)` | |
| `= -(ln 4.5 – ln 5)` | |
| `= -ln {:9/10` | |
| `= ln {:10/9` |
Use the substitution `u = e^(3x)` to evaluate `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`. (3 marks)
--- 8 WORK AREA LINES (style=lined) ---
`1/3 (tan^(-1)e\ – pi/4)`
`text(Let)\ \ u = e^(3x)`
| `(du)/(dx)` | `= 3e^(3x)` |
| `:.dx` | `= (du)/(3e^(3x))` |
| `text(When)` | `\ x = 1/3,` | `\ u = e^(3 xx 1/3) = e` |
| `\ x = 0,` | `\ u = e^0 = 1` |
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`
`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`
`= 1/3 int_1^e 1/(u^2 + 1)\ du`
`= 1/3 [tan^(-1)u]_1^e`
`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`
`= 1/3 (tan^(-1)e\ – pi/4)`