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Calculus, EXT1 C2 EQ-Bank 30

Solve \( \displaystyle\int_0^1 \dfrac{1}{(1+x^2)^\tfrac{3}{2}}\, dx \)  using the substitution  \(x=\tan \theta\).   (3 marks)

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\(\dfrac{1}{\sqrt{2}}\)

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\(\text {Let} \ \ x=\tan \theta\)

\(\dfrac{d x}{d \theta}=\sec ^2 \theta \ \Rightarrow \ d x=\sec ^2 \theta \, d \theta\)

\(\text{Also,} \ \ 1+x^2=1+\tan ^2 \theta=\sec ^2 \theta\)

\(\left(1+x^2\right)^{\tfrac{3}{2}}=(\sec ^2 \theta)^{\tfrac{3}{2}}=\sec ^3 \theta\)
 

\(\text{Adjust limits:}\)

\(x=0 \Rightarrow \theta=0, \ \ x=1 \Rightarrow \theta=\dfrac{\pi}{4}\)

\(\displaystyle\int_0^1 \dfrac{1}{\left(1+x^2\right)^{\tfrac{3}{2}}} \, d x\) \(=\displaystyle\int_0^{\tfrac{\pi}{4}} \frac{\sec ^2 \theta}{\sec ^3 \theta} \, d \theta\)
  \(=\displaystyle \int_0^{\tfrac{\pi}{4}} \cos \theta \, d \theta\)
  \(=\Big[\sin \theta\Big]_0^{\tfrac{\pi}{4}}\)
  \(=\sin \dfrac{\pi}{4}-\sin 0\)
  \(=\dfrac{1}{\sqrt{2}}\)

Calculus, EXT1 C2 2025 SPEC2 7 MC

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

  1. \(\displaystyle\int_0^{\tfrac{\pi}{2}} u \sqrt{\frac{1-u}{1+u}}\ d u\)
  2. \(\displaystyle\int_0^1\left(1+\frac{1}{1+u}\right) d u\)
  3. \(\displaystyle\int_0^{\tfrac{\pi}{2}}\left(1-\frac{1}{1+u}\right) d u\)
  4. \(\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) d u\)
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\(D\)

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\(u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta\)

\(\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0\)

\(\text{When}\ \  \theta=0, u=1\)

\(I\) \(=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta\)
  \(=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta\)
  \(=\displaystyle \int_1^0-\frac{u}{1+u} \, d u\)
  \(=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u\)
  \(=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u\)

 

\(\Rightarrow D\)

Calculus, EXT1 C2 2025 HSC 3 MC

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?

  1. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec} \theta \, d \theta\)
  2. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \,  d \theta\)
  3. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec}^2 \theta \, d \theta\)
  4. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec ^2 \theta \, d \theta\)
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\(\Rightarrow B\)

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\(x=5\, \sin \theta\)

\(\dfrac{dx}{d \theta}=5\, \cos \theta \ \Rightarrow \ dx=5\, \cos \theta \, d \theta\)

\(\text{When} \ \ x=\dfrac{5}{2} \ \Rightarrow \ \sin\, \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text{When} \ \ x=-\dfrac{5}{2} \ \Rightarrow \ \sin \theta=-\dfrac{1}{2} \ \Rightarrow \ \theta=-\dfrac{\pi}{6}\)

\(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\) \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\left(\dfrac{1}{25-25\, \sin ^2 \theta}\right) \cdot 5\  \cos \theta \, d \theta\)
  \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{5\, \cos \theta}{25\, \cos ^2 \theta} \,  d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{1}{\cos \theta} \, d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \, d \theta\)

\(\Rightarrow B\)

Calculus, EXT1 C2 EQ-Bank 28

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)

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\(\text{See worked solutions}\)

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\(x=\cos 2 \theta\)

\(\dfrac{dx}{d \theta}=-2 \sin 2 \theta \ \Rightarrow\ \ dx=-2 \sin 2 \theta\, d \theta\)

\(\displaystyle \int \sqrt{\frac{1-x}{1+x}} \, dx\) \(=\displaystyle\int \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}} \times -2 \sin 2 \theta\, d \theta\)
  \(=\displaystyle\int \sqrt{\frac{1-\left(2 \cos ^2 \theta-1\right)}{1+\left(2 \cos ^2 \theta-1\right)}} \times -4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int \sqrt{\frac{2\left(1-\cos ^2 \theta\right)}{2 \cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int\sqrt{\dfrac{\sin ^2 \theta}{\cos ^2 \theta}} \times-4 \sin \theta \, \cos \theta \, d \theta\)
  \(=\displaystyle \int-4 \sin ^2 \theta \, d \theta\)
  \(=-4 \displaystyle \int \frac{1-\cos 2 \theta}{2} \,d \theta\)
  \(=-2 \displaystyle \int 1-\cos 2 \theta \, d \theta\)
  \(=-2 \displaystyle \int 1\, d \theta+2 \int \cos 2 \theta \, d \theta\)
  \(=-2 \theta+\sin 2 \theta+c\)
  \(=-\cos ^{-1} x+\sqrt{1-\cos ^2 2 \theta}+c \quad \text {(note:}\ \  x=\cos 2 \theta \Rightarrow 2 \theta=\cos ^{-1} x \text{)}\)
  \(=-\cos ^{-1} x+\sqrt{1-x^2}+c\)

Calculus, EXT1 C2 EQ-Bank 20

Use the substitution  \(u=\cos\,x\)  to evaluate

\(\displaystyle {\int}_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin x\, dx\).   (3 marks)

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\(1-\dfrac{1}{e}\)

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\(u=\cos x\)

\(\dfrac{d u}{d x}=-\sin\,x\  \Rightarrow\  du=-\sin\,x\,dx\)

\(\text {When} \ \ x=\pi, u=-1\)

\(\text{When} \ \ x=\dfrac{\pi}{2}, u=0\)

\(\displaystyle \int_{\frac{\pi}{2}}^\pi e^{\cos\,x} \sin\,x\,d x\) \(=\displaystyle -\int_0^{-1} e^u\,d u\)
  \(=\left[-e^u\right]_0^{-1}\)
  \(=-e^{-1}+e^0\)
  \(=1-\dfrac{1}{e}\)

Calculus, EXT1 C2 2024 HSC 13d

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)

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\(\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

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\(u=e^x+2 e^{-x} \ \Rightarrow \ u^2=\left(e^x+2 e^{-x}\right)^2=e^{2 x}+4+4 e^{-2 x}\)

\(\dfrac{du}{dx}=e^x-2 e^{-x} \ \Rightarrow \ du=\left(e^x-2 e^{-x}\right)\, d x\)

Mean mark 55%.
  \(\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}}\, d x\)
    \(=\displaystyle \int \frac{e^{3 x}-2 e^x}{4+8 e^{2 x}+e^{4 x}} \times \frac{e^{-2 x}}{e^{-2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4 e^{-2 x}+8+e^{2 x}}\, d x\)
    \(=\displaystyle \int \frac{e^x-2 e^{-x}}{4+\left(e^{2 x}+4+4 e^{-2 x}\right)}\, d x\)
    \(=\displaystyle \int \frac{1}{4+u^2}\, d u\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{u}{2}\right)+c\)
    \(=\displaystyle \frac{1}{2} \tan ^{-1}\left(\frac{e^x+2 e^{-x}}{2}\right)+c\)

Calculus, EXT1 C2 2020 HSC 13a

  1. Find  `d/(d theta) (sin^3 theta)`.  (1 mark)

  2. Use the substitution  `x = tan theta`  to evaluate  `int_0^1 (x^2)/(1 + x^2)^(5/2)\ dx`.  (4 marks)

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  1. `3 cos theta sin^2 theta`
  2. `sqrt2/12`
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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`

Calculus, EXT1 C2 2019 HSC 13a

Use the substitution  `u = cos^2 x`  to evaluate  `int_0^(pi/4) (sin 2x)/(4 + cos^2 x)\ dx`.  (3 marks)

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`ln {:10/9`

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`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`

Calculus, EXT1 C2 2013 HSC 11f

Use the substitution  `u = e^(3x)`  to evaluate  `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`.   (3 marks)

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`1/3 (tan^(-1)e\ – pi/4)`

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 MARKER’S COMMENT: Many students did not calculate in radians and incorrectly got an answer of 8.2. BE CAREFUL!
Note that converting your answer to 0.14 is also correct but not required.

`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)`