Substitution and Harder Integration

Calculus, EXT2 C1 2025 HSC 13d

Evaluate \(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\), by first using the substitution \(u=\dfrac{\pi}{2}-x\). (4 marks)

--- 15 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution

\(\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{u}{1+\sin u+\cos u} \, du\)

\(\text{Let} \ \ u=\dfrac{\pi}{2}-x \ \ \Rightarrow\ \ \dfrac{du}{dx}=-1 \ \ \Rightarrow\ \ du=-dx\)

\(\text{Limits:} \ \ u=\dfrac{\pi}{2}\ \ \Rightarrow\ \ x=0, \ \ u=0\ \ \Rightarrow\ \ x=\dfrac{\pi}{2}\)

\(I\)	\(=-\displaystyle \large{\int_{\small{\dfrac{\pi}{2}}}^0}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\sin \left(\dfrac{\pi}{2}-x\right)+\cos \left(\dfrac{\pi}{2}-x\right)}\,dx\)
	\(=\displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{\dfrac{\pi}{2}-x}{1+\cos x+\sin x}\, d x\)

\(\text{Add \(I\) (swap variable from \(x\) to \(u\)) to original integral:}\)

\(2I=\dfrac{\pi}{2} \displaystyle \large{\int_0^{\small{\dfrac{\pi}{2}}}}\)\(\dfrac{1}{1+\sin u+\cos u}\, d u\)

\(\text{Substitute} \ \ t=\tan \left(\frac{u}{2}\right), \ \sin u=\dfrac{2t}{1+t^2}, \ \cos u=\dfrac{1-t^2}{1+t^2}\)

\(d t=\dfrac{1}{2} \sec ^2\left(\frac{u}{2}\right)\, du \ \ \Rightarrow\ \ du=\dfrac{2}{1+\tan ^2\left(\frac{u}{2}\right)}\, dt=\dfrac{2}{1+t^2}\, d t\)

\(\text{Limits:} \ \ n=\dfrac{\pi}{2}\ \Rightarrow \ t=1, \ n=0 \ \Rightarrow \ t=0\)

\(2I\)	\(=\dfrac{\pi}{2} \displaystyle \int_0^1 \dfrac{1}{1+\frac{2 t}{1+t^2}+\frac{1-t^2}{1+t^2}} \times \frac{2}{1+t^2}\,d t\)
\(I\)	\(=\displaystyle\frac{\pi}{4} \int_0^1 \frac{2}{1+t^2+2 t+1-t^2}\, d t\)
	\(=\displaystyle \frac{\pi}{4} \int_0^1 \frac{1}{1+t}\, d t\)
	\(=\dfrac{\pi}{4}\Bigl[\ln (1+t)\Bigr]_0^1\)
	\(=\dfrac{\pi}{4}(\ln 2-\ln 1)\)
	\(=\dfrac{\pi \, \ln 2}{4}\)

Calculus, EXT2 C1 2025 HSC 13a

It is given that \(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\, dx\).

Show that \(\displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=k A\), where \(k\) and \(m\) are constants. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

Show Answers Only

\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\)	\(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
	\(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
	\(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Show Worked Solution

\(A=\displaystyle \int_2^4 \frac{e^x}{x-1}\,d x\)

\(\text{Let} \ \ u=m-x\)

\(\dfrac{d u}{d x}=-1 \ \ \Rightarrow \ \ du=-d x\)

♦♦ Mean mark 33%.

\(\text{When} \ \ x=4, \ u=m-4\)

\(\text{When} \ \ x=2, \ u=m-2\)

\(A\)	\(=-\displaystyle\int_{m-2}^{m-4} \frac{e^{m-u}}{m-u-1}\, du\)
	\(=-e^m \displaystyle \int_{m-2}^{m-4} \frac{e^{-u}}{m-u-1}\, du\)
	\(=-e^m \times \left[\displaystyle \int_{m-4}^{m-2} \frac{e^{-u}}{u-m+1}\, du\right]\)

\(\Rightarrow \displaystyle \int_{m-4}^{m-2} \frac{e^{-x}}{x-m+1}\, d x=kA \ \ (\text{where}\ \ k=-e^{-m})\)

Calculus, EXT2 C1 2024 HSC 15d

Using a suitable substitution, find \(\displaystyle\int \dfrac{2 x^2}{\sqrt{2 x-x^2}}\, d x\). (3 marks)

--- 12 WORK AREA LINES (style=lined) ---

Show Answers Only

\(I=3 \sin ^{-1}(x-1)-4 \sqrt{2x-x^2}-(x-1) \sqrt{2x-x^2}+c\)

Show Worked Solution

	\(\displaystyle \int \dfrac{2x^2}{\sqrt{2x-x^2}}\, dx\)	\(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(1-2 x+x^2)}}\, dx\)
		\(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(x-1)^2}} \, dx\)

♦ Mean mark 49%.

\(\text {Let} \ \ x-1=\sin \theta \ \Rightarrow \ x=1+\sin \theta\)

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

	\(I\)	\(=\displaystyle \int \frac{2(1+\sin \theta)^2}{\sqrt{1-\sin ^2 \theta}} \cdot \cos \theta \, d \theta\)
		\(=2 \displaystyle \int \frac{1+2 \sin \theta+\sin ^2 \theta}{\cos \theta} \cdot \cos \theta \, d \theta\)
		\(=2 \displaystyle \int 1+2 \sin \theta+\frac{1}{2}(1-\cos (2 \theta)) \, d \theta\)
		\(=\displaystyle \int 2+4 \sin \theta+1-\cos (2 \theta) \, d \theta\)
		\(=\displaystyle \int 3+4 \sin \theta-\cos (2 \theta) \, d \theta\)
		\(=3 \theta-4 \cos \theta-\dfrac{1}{2} \sin (2 \theta)+c\)
		\(=3 \theta-4 \cos \theta-\sin \theta \cos \theta+c\)

\(\Rightarrow \cos \theta=\sqrt{1^2-(x-1)^2}=\sqrt{2x-x^2}\)

\(\therefore I=3 \sin ^{-1}(x-1)-4 \sqrt{2x-x^2}-(x-1) \sqrt{2x-x^2}+c\)

Calculus, EXT2 C1 2022 SPEC2 7 MC

Using the substitution `u=1+e^x, \int_0^{\log _e 2} \frac{1}{1+e^x}dx` can be expressed as

`\int_0^{\log _e 2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
`\int_2^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
`\int_1^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
`\int_2^3\left(\frac{1}{u-1}-\frac{1}{u}\right) du`

Show Answers Only

`D`

Show Worked Solution

`u=1+e^x\ \ =>\ \ \frac{du}{dx}=e^x\ \ =>\ \ dx=\frac{du}{e^x}=\frac{du}{u-1}`

`text{When}\ \ x=0,\ \ u=2`

`text{When}\ \ x=\log_e 2,\ \ u=1+e^{\log_e 2} = 3`

`\int_0^{\log _e 2} \frac{1}{1+e^x}dx = \int_2^3\left(\frac{1}{u}*\frac{1}{u-1}\right) du= \int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`

`=>D`

Calculus, EXT2 C1 2022 SPEC1 9

Given that `f^{\prime}(x)=\frac{\cos (2 x)}{\sin ^3(2 x)}` and `f((pi)/(8))=(3)/(4)`, find `f(x)`. (4 marks)

--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

`f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

Show Worked Solution

`text{Let}\ \ u=sin(2x)\ \ =>\ \ {du}/{dx}=2cos(2x)`

`\int \frac{\cos (2 x)}{\sin ^3(2 x)} d x`	`=\frac{1}{2} \int \frac{d u}{u^3}`
	`= -\frac{1}{4} u^{-2}+c`
	`=-\frac{1}{4} \ xx \frac{1}{\sin ^2(2 x)}+c`

`text{When}\ \ x = pi/8:`

`-\frac{1}{4} \cdot \frac{1}{(\frac{1}{sqrt{2}})^2}+c`	`=\frac{3}{4}`
`-\frac{1}{4} \cdot 2+c`	`=\frac{3}{4}`
`\Rightarrow c`	`=\frac{3}{4}+\frac{2}{4}=5/4`

`:. \ f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

♦ Mean mark 50%.

Calculus, EXT2 C1 2023 HSC 13a

Find \({\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\). (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

\(I=\sqrt{5-4x-x^2} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

Show Worked Solution

\(I\)	\(={\displaystyle \int \frac{1-x}{\sqrt{5-4 x-x^2}}\ dx}\)
	\(= \dfrac{1}{2} {\displaystyle \int \frac{2-2x}{\sqrt{5-4 x-x^2}}\ dx}\)
	\(=\dfrac{1}{2} {\displaystyle \int \frac{-4-2x}{\sqrt{5-4 x-x^2}}\ dx} + \dfrac{1}{2} {\displaystyle \int \frac{6}{\sqrt{5-4 x-x^2}}\ dx}\)

\(\text{Let}\ \ u=5-4x-x^2 \)

\( \dfrac{du}{dx}=-4-2x\ \ \Rightarrow\ \ du=-4-2x\ dx \)

\(I\)	\(= \dfrac{1}{2} {\displaystyle \int u^{-\frac{1}{2}}\ du} + {\displaystyle 3 \int \frac{1}{\sqrt{9-(x+2)^2}}\ dx}\)
	\(= u^{\frac{1}{2}} + 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)
	\(= \sqrt{5-4x-x^2}+ 3 \sin^{-1} \big{(} \dfrac{x+2}{3} \big{)} + c \)

Calculus, EXT2 C1 2022 HSC 11f

Using the substitution `t=tan\ x/2`, find

`int(dx)/(1+cos x-sin x)` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`-lnabs(1-tan(x/2))+C`

Show Worked Solution

`text(Let)\ \ t = tan\ x/2, \ cos\ x = (1-t^2)/(1 + t^2), \ sin\ x=(2t)/(1+t^2)`

`dt = 1/2 sec^2\ x/2\ dx \ => \ d x = (2\ dt)/(sec^2\ x/2) = 2/(1 + t^2)\ dt`

`text{I}`	`= int(dx)/(1+cos x-sin x)`
	`=int 1/(1+(1-t^2)/(1 + t^2)-(2t)/(1 + t^2)) *2/(1 + t^2)\ dt`
	`=int 2/(1+t^2+1-t^2-2t)\ dt`
	`=int 1/(1-t)\ dt`
	`=-ln abs(1-t)+C`
	`=-lnabs(1-tan(x/2))+C`

Calculus, EXT2 C1 2022 HSC 5 MC

If `int_(a)^(x)f(t)dt=g(x)`, which of the following is a primitive of `f(x)g(x)` ?

`(1)/(2)[f(x)]^(2)`
`(1)/(2)[f^(')(x)]^(2)`
`(1)/(2)[g(x)]^(2)`
`(1)/(2)[g^(')(x)]^(2)`

Show Answers Only

`C`

Show Worked Solution

`int_a^x f(t)\ dt`	`=g(x)`
`d/dx [int_a^xf(t)\ dt]`	`=g^{′}(x)`
`f(x)`	`=g^{′}(x)`
`f(x)*g(x)`	`=g^{′}(x)g(x)`
`int f(x)*g(x)\ dx`	`=int g^{′}(x)g(x)\ dx`
	`=1/2[g(x)]^2+C`

`=>C`

COMMENT: Simple examples can illustrate the logic of line 3 of the worked solution.

Calculus, EXT2 C1 2022 HSC 4 MC

Of the following expressions, which one need NOT contain a term involving a logarithm in its anti-derivative?

`(x+2)/(x^(2)+4x+5)`
`(x+2)/(x^(2)-4x-5)`
`(x-1)/(x^(3)-x^(2)+x-1)`
`(x+1)/(x^(3)-x^(2)+x-1)`

Show Answers Only

`C`

Show Worked Solution

`text{Consider the denominator of}\ C:`

`x^(3)-x^(2)+x-1`	`=x^2(x-1)+(x-1)`
	`=(x^2+1)(x-1)`

`(x-1)/(x^(3)-x^(2)+x-1)=(x-1)/((x^2+1)(x-1))=1/(x^2+1)`

`int 1/(x^2+1)\ dx=tan^(-1)(x)+c`

`=>C`

Calculus, EXT2 C1 2021 HSC 13b

Use an appropriate substitution to evaluate `int_(sqrt10)^(sqrt13) x^3 sqrt{x^2-9}\ dx`. (3 marks)

--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

`136/5`

Show Worked Solution

`int_(sqrt10)^(sqrt13) x^3 sqrt{x^2-9} dx`

`text{Let} \ \ u = x^2-9 \ => \ x^2 = u + 9`

`(du)/dx = 2x \ => \ du = 2x\ dx`

`text{If} \ \ x = sqrt13 \ , \ u =4`

`text{If} \ \ x = sqrt10 \ , \ u = 1`

`int_(sqrt10)^(sqrt13) x^3 sqrt{x^2-9}\ dx`	`= int_(1)^(4) (u + 9) sqrtu * 1/2\ du`
	`= 1/2 int_(1)^(4) u^{3/2} + 9 u^{1/2}\ du`
	`= 1/2 [ 2/5 u^{5/2} + 9 * 2/3 u^{3/2}]_(1)^(4)`
	`= 1/2 [(2/5 * 4^{5/2} + 6 * 4^{3/2})-(2/5 + 6)]`
	`= 1/2 (64/5 + 48-32/5)`
	`= 136/5`

Calculus, EXT2 C2 2020 SPEC2 11 MC

With a suitable substitution `int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) - 3 tan(x) + 1)\ dx` can be expressed as

`int_1^(1/sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
`int_1^(sqrt3) (1/(u - 2) - 1/(u - 1))\ du`
`int_1^(sqrt3) (1/(u - 1) - 1/(u - 2))\ du`
`int_(pi/4)^(pi/3) (1/(3(u - 1)) - 1/(3(u + 2)))\ du`

Show Answers Only

`B`

Show Worked Solution

`text(Let)\ \ u = tan(x)`

`(du)/(dx) = sec^2 (x) \ => \ du = sec^2(x)\ dx`

`text(When)\ `	`x = pi/3,\ u = sqrt3`
	`x = pi/4,\ u = 1`

`int_(pi/4)^(pi/3) (sec^2(x))/(sec^2(x) – 3 tan(x) + 1)\ dx`	`= int_1^(sqrt3)\ 1/(u^2 + 1 – 3u + 1)\ du`
	`= int_1^(sqrt3) 1/(u^2 – 3u + 2)\ du`
	`= int_1^(sqrt3) 1/((u – 2)(u – 1))\ du`
	`= int_1^(sqrt3) 1/(u – 2) – 1/(u – 1)\ du`

`=>B`

Calculus, EXT2 C1 2020 SPEC1 2

Evaluate `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`(8 sqrt 2)/3 – 10/3`

Show Worked Solution

`text(Let)\ \ u`	`= 1 – x \ => \ x = 1 – u`
`(du)/(dx)`	`= -1 \ => \ dx = -du`

`text(When)\ \ x`	`= 0,\ u = 1`
`x`	`= -1,\ u = 2`

`int_(-1)^0 (1 + x)/sqrt(1 – x)\ dx`	`= -int_2^1 (2 – u)/sqrt u\ du`
	`= int_1^2 2u^(-1/2) – u^(1/2)\ du`
	`= [4u^(1/2) – 2/3u^(3/2)]_1^2`
	`= 4 sqrt 2 – (4 sqrt 2)/3 – (4 – 2/3)`
	`= (8 sqrt 2)/3 – 10/3`

Calculus, EXT2 C1 2020 HSC 10 MC

Which of the following is equal to `int_0^(2a) f(x)\ dx`?

`int_0^a f(x) - f(2a - x)\ dx`
`int_0^a f(x) + f(2a -x)\ dx`
`2 int_0^a f(x - a)\ dx`
`int_0^a frac{1}{2} f(2x)\ dx`

Show Answers Only

`B`

Show Worked Solution

`int_0^(2a) f(x)\ dx – int_0^a f(x)\ dx + int_a^(2a) f(x)\ dx`

♦ Mean mark 54%.

`text(Let)\ \ x = 2a – u\ \ => \ u = 2a – x`

`frac{du}{dx} = -1 \ \ => \ du = -dx`

`text{When}`	`x = a,`	`\ u = a`
	`x= 2a,`	`\ u = 0`

`int_0^(2a) f(x)\ dx= int_0^a f(x)\ dx – int_a^0 f(2a – u)\ du`

`text(Use substitution for)\ \ int_a^0 f(2a – u)\ du`

`text(Let)\ \ 2a – u = 2a-x \ \ => \ x=u`

`dx/(du) = 1 \ => \ du=dx`

`text{When}`	`u = a,`	`\ x = a`
	`u= 0,`	`\ x = 0`

`:. int_0^(2a) f(x)\ dx`	`= int_0^a f(x)\ dx – int_a^0 f(2a – x)\ dx`
	`= int_0^a f(x)\ dx + int_0^a f(2a -x)\ dx`
	`= int_0^a f(x) + f(2a – x)\ dx`

`=> \ B`

Calculus, EXT2 C1 2019 NHT 4

Evaluate `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`log_e ((4)/(3))`

Show Worked Solution

`text(Let)\ \ u = log_e x`

`(du)/(dx) = (1)/(x) \ => \ du = (1)/(x) dx`

`text(When) \ \ x = e^4 \ => \ u = 4`

`text(When) \ \ x = e^3 \ => \ u = 3`

`int_(e^3) ^(e^4) (1)/(x log_e (x))`	`= int_3 ^4 (1)/(u)\ du`
	`= [ log_e u]_3 ^4`
	`= log_e 4 – log_e 3`
	`= log_e ((4)/(3))`

Calculus, EXT2 C1 2005 HSC 1a

Find `int(cos theta)/(sin^5 theta) d theta` (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`(-1)/(4sin^4 theta ) + C`

Show Worked Solution

`text(Let) \ u`	`= sin theta `
`(du)/(d theta)`	`=cos theta => d u = cos theta \ d theta`

`int(cos theta)/(sin^5 theta) d theta`	`= int u^-5 d u`
	`=(-1)/(4) u ^-4 + C`
	`=(-1)/(4sin^4 theta ) + C`

Calculus, EXT2 C1 2019 HSC 15a

Show that

`qquad int_(-a)^a (f(x))/(f(x) + f(-x))\ dx = int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, or otherwise, evaluate

`qquad int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ text{(See Worked Solutions)}`
`text(See Worked Solutions)`

Show Worked Solution

i. `text(Let)\ \ u = -x \ => \ du = -dx`

`text(When)\ \ x=a, \ u=-a`

`text(When)\ \ x=-a, \ u=a`

`int_(-a)^a (f(x))/(f(x) + f(-x))\ dx`	`= -int_a^(-a) (f(-u))/(f(-u) + f(u))\ du`
	`= int_(-a)^a (f(-u))/(f(u) + f(-u))\ du`
	`= int_(-a)^a (f(-x))/(f(x) + f(-x))\ dx`

ii. `text(Let)\ \ I = int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx`

`2I`	`= int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx + int_(-1)^1 (e^(-x))/(e^x + e^(-x))\ dx`
`2I`	`= int_(-1)^1 (e^x + e^(-x))/(e^x + e^(-x))\ dx`
`2I`	`= [x]_(-1)^1`
`2I`	`= 2`

`:. int_(-1)^1 (e^x)/(e^x + e^(-x))\ dx = 1`

Calculus, EXT2 C1 2017 HSC 7 MC

It is given that `f(x)` is a non-zero even function and `g(x)` is a non-zero odd function.

Which expression is equal to `int_(−a)^a f(x) + g(x)\ dx`?

`2 int_0^a f(x)\ dx`
`2 int_0^a g(x)\ dx`
`int_(−a)^a g(x)\ dx`
`2int_0^a f(x) + g(x)\ dx`

Show Answers Only

`A`

Show Worked Solution

`int_(−a)^a f(x) + g(x)\ dx`	`= int_(−a)^a f(x)\ dx + int_(−a)^a g(x)\ dx`
	`= 2 int_0^a f(x)\ dx + 0`

`=> A`

Calculus, EXT2 C1 2017 HSC 11f

Using the substitution `x = sin^2 theta`, or otherwise, evaluate `int_0^(1/2) sqrt(x/(1 - x))\ dx`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`pi/4 – 1/2`

Show Worked Solution

`x = sin^2 theta`

`dx = 2 sin theta cos theta\ d theta`

`text(If)\ \ x = 1/2, sin theta = 1/sqrt 2, \ theta = pi/4`

`text(If)\ \ x = 0, \ theta = 0`

`int_0^(1/2) sqrt(x/(1 – x))\ dx`	`= int_0^(pi/4) sqrt ((sin^2 theta)/(1 – sin^2 theta)) xx 2 sin theta cos theta\ d theta`
	`= int_0^(pi/4) (sin theta)/(cos theta) xx 2 sin theta cos theta\ d theta`
	`= int_0^(pi/4) 2 sin^2 theta\ d theta`
	`= int_0^(pi/4) 1 – cos 2 theta\ d theta`
	`= [theta – 1/2 sin 2 theta]_0^(pi/4)`
	`= (pi/4 – 1/2 sin {:pi/2) – (0 – 0)`
	`= pi/4 – 1/2`

Calculus, EXT2 C1 2007 HSC 8a

Using a suitable substitution, show that

`int_0^a f(x)\ dx = int_0^a f(a - x)\ dx.` (1 mark)

--- 6 WORK AREA LINES (style=lined) ---
A function `f(x)` has the property that `f(x) + f(a - x) = f(a).`

Using part (i), or otherwise, show that

`int_0^a f(x)\ dx = a/2\ f(a).` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a. `text(Show)\ \ int_0^a f(x)\ dx = int_0^a f(a – x)\ dx.`

`text(Let)\ \ x = a – u,\ \ dx = -du`

`text(When)\ \ x = 0,\ \ u = a`

`text(When)\ \ x = a,\ \ u = 0`

`:. int_0^a f(x)\ dx`	`=int_a^0 f(a – u) (-du)`
	`=int_0^a f(a – u)\ du`
	`=int_0^a f(a – x)\ dx\ \ text{.. as required}`

MARKER’S COMMENT: Integrating `f(a)` was poorly done in part (ii). Pay careful attention to this in the Worked Solution.

b. `f(x) = f(a) – f(a – x)`

`int_0^a f(x)\ dx`	`=int_0^a [f(a) – f(a – x)]\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(a – x)\ dx`
	`=int_0^a f(a)\ dx – int_0^a f(x)\ dx`
`2 int_0^a f(x)\ dx`	`=int_0^a f(a)\ dx`
	`=[f(a) xx x]_0^a`
`int_0^a f(a)\ dx`	`=(f(a))/2 (a – 0)`
	`=a/2\ f(a)`

Calculus, EXT2 C1 2007 HSC 5c

Write `(x - 1) (5 - x)` in the form `b^2 - (x - a)^2`, where `a` and `b` are real numbers. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Using the values of `a` and `b` found in part (i) and making the substitution `x - a = b sin theta`, or otherwise, evaluate

` int_1^5 sqrt ((x - 1) (5 - x))\ dx.` (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `2^2 – (x – 3)^2`

b. `2 pi`

Show Worked Solution

a. `(x – 1) (5 – x)`	`= 5x – x^2 -5+ x`
	`= -(x^2 – 6x + 5)`
	`= 4 – (x – 3)^2`
	`= 2^2 – (x – 3)^2`

b.	`text(Let)\ \ x – 3`	`= 2 sin theta`
	`dx`	`= 2 cos theta\ d theta`

`text(When)\ \ x = 1,\ \ theta = -pi/2`

`text(When)\ \ x = 5,\ \ theta = pi/2`

`:.int_1^5 sqrt (4 – (x – 3)^2)\ dx`

`=int_((-pi)/2)^(pi/2) sqrt ((4 – 4 sin^2 theta))*2 cos theta\ d theta`

`=4 int_((-pi)/2)^(pi/2) sqrt(cos^2 theta) * cos theta\ d theta`

`=4 int_((-pi)/2)^(pi/2) cos^2 theta\ d theta`

`=2 int_((-pi)/2)^(pi/2) (1 + cos 2 theta)\ d theta`

`=2[theta + (sin 2 theta)/2]_((-pi)/2)^(pi/2)`

`=2[(pi/2 + 0) – ((-pi)/2 – 0)]`

`=2 pi`

Calculus, EXT2 C1 2012 HSC 10 MC

Without evaluating the integrals, which one of the following integrals is greater than zero?

`int_(−pi/2)^(pi/2) x/(2 + cos x)\ dx`
`int_(−pi)^pi x^3 sin x\ dx`
`int_(−1)^1 (e^(−x^2) − 1)\ dx`
`int_(−2)^2 tan^(−1)(x^3)\ dx`

Show Answers Only

`B`

Show Worked Solution

`text{Consider (A) and (D)}`

`f(x)=-f(-x)\ \ =>\ text(ODD functions where)`

`int_(−a)^a f(x)\ dx = 0`

`text{Consider (C)}`

`e^(−x^2)<1\ \ text(for all)\ x\ \ => e^(−x^2) − 1<0`

`:. text(Its graph is below the)\ x text(-axis and any integral)`

`text(will be negative)`

`text{Consider (B)}`

`text{(B)}\ text(is an even function where,)`

`x^3 sinx>=0\ \ text(for)\ \ \ -pi<=x<=pi`

`=>B`

Calculus, EXT2 C1 2007 HSC 1d

Evaluate `int_0^(3/4) x/sqrt (1 - x)\ dx.` (4 marks)

Show Answers Only

`5/12`

Show Worked Solution

`text(Let)\ \ u = 1 – x,\ \ du = -dx,\ \ x = 1 – u`

`text(When)\ \ x = 0,\ \ u = 1`

`text(When)\ \ x = 3/4,\ \ u = 1/4`

`int_0^(3/4) (x\ dx)/sqrt (1 – x)`	`=int_1^(1/4) ((1 – u)(-du))/sqrt u`
	`=int_(1/4)^1 (u^(-1/2) – u^(1/2)) du`
	`=[2 u^(1/2) – 2/3 u^(3/2)]_(1/4)^1`
	`=[(2 – 2/3) – (1 – 1/12)]`
	`=5/12`

Calculus, EXT2 C1 2014 HSC 10 MC

Which integral is necessarily equal to `int_(−a)^a f(x)\ dx`?

`int_0^a f(x) − f(−x)\ dx`
`int_0^a f(x) − f(a − x)\ dx`
`int_0^a f(x − a) + f(−x)\ dx`
`int_0^a f(x − a) + f(a − x)\ dx`

Show Answers Only

`D`

Show Worked Solution

`int_(−a)^a f(x)\ dx= int_0^a f(x)\ dx + int_(−a)^0 f(x)\ dx`

`text(Consider)\ \ int_0^a f(x)\ dx`

♦♦ Mean mark 29%.

`text(Let)\ u = a − x\ \ => x = a − u\ \ text(and)\ \ dx = −du`

`int_0^a f(x)\ dx`	`= int_0^a f(a − u) − du`
	`=int_a^0 f(a-u)\ du`
	`= int_0^a f(a − x)\ dx`

`text(Consider)\ \ int_(−a)^0 f(x)\ dx`

`text(Let)\ u = x + a\ \ => x = u − a\ \ text(and)\ \ dx = du`

`int_(−a)^0 f(x)\ dx`	`= int_0^a f(u − a)\ du`
	`= int_0^a f(x − a)\ dx`
`:.int_(−a)^a f(x)\ dx`	`= int_0^a f(x − a)+ f(a − x)\ dx`

`=> D`

Calculus, EXT2 C1 2006 HSC 1a

Find `int x/sqrt (9 - 4x^2)\ dx.` (2 marks)

Show Answers Only

`- 1/4 sqrt (9 – 4x^2)+c`

Show Worked Solution

`int x/sqrt (9 – 4x^2)\ dx`	`=-1/8 int (-8x)/sqrt (9 – 4x^2)\ dx`
	`=-1/4 sqrt (9 – 4x^2) + c`

Calculus, EXT2 C1 2009 HSC 1e

Evaluate `int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx.` (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`sqrt 2 – (2 sqrt 3)/3`

Show Worked Solution

`text(Let)\ \ x = tan theta,\ \ \ \ dx = sec^2 theta\ d theta`

`text(When)\ \ x = 1,\ \ theta = pi/4`

`text(When)\ \ x = sqrt 3,\ \ theta = pi/3`

`int_1^sqrt 3 1/(x^2 sqrt (1 + x^2))\ dx`	`= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sqrt (1 + tan^2 theta))\ d theta`
	`= int_(pi/4)^(pi/3) (sec^2 theta)/(tan^2 theta sec theta)\ d theta`
	`= int_(pi/4)^(pi/3) (sec theta)/(tan^2 theta)\ d theta`
	`= int_(pi/4)^(pi/3) (cos^2 theta)/(cos theta sin^2 theta)\ d theta`
	`= int_(pi/4)^(pi/3) (cos theta\ d theta)/(sin^2 theta)`
	`= [-1/(sin theta)]_(pi/4)^(pi/3)`
	`= (-2/sqrt 3 + sqrt 2)`
	`= sqrt 2 – (2 sqrt 3)/3`

Calculus, EXT2 C1 2009 HSC 1a

Find `int (ln x)/x\ dx.` (2 marks)

Show Answers Only

`((ln x)^2)/2 + c`

Show Worked Solution

`text(Let)\ \ u=lnx,\ \ \ du=1/x\ dx`

`int (ln x)/x \ dx`	`=int u\ du`
	`=1/2 u^2 +c`
	`=1/2 (ln x)^2 +c`

Calculus, EXT2 C1 2010 HSC 1e

Find `int (dx)/(1 + sqrtx)`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`2sqrtx − 2log_e\ (1 + sqrtx) + c_1\ \ text(or)`

`2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

Show Worked Solution

`text(Solution 1)`

`text(Let)\ \ u = sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2u\ du`
`int (dx)/(1 + sqrtx)`	`=int (2u\ du)/(1 + u)`
	`=int((2+2u-2)/(1+u))\ du`
	`=int (2 − 2/(1 + u))\ du`
	`=2u − 2log_e(1 + u) + c_1`
	`=2sqrtx − 2log_e\ (1 + sqrtx) + c_1`

`text(Alternative Solution)`

`text(Let)\ \ u = 1+sqrtx,\ \ du = 1/(2sqrtx)\ dx,\ \ dx = 2(u-1)\ du`
`int (dx)/(1 + sqrtx)`	`=2int (u-1)/u\ du`
	`=2 int(1 − 1/u)du`
	`=2u − 2log_e\ \|\ u\ \| + c_2`
	`=2+2sqrtx − 2log_e\ (1 + sqrtx) + c_2`

`text(NB. These solutions are equivalent by making)\ \ c_1=c_2+2`

Calculus, EXT2 C1 2010 HSC 1b

Evaluate `int_0^(pi/4) tan\ x\ dx`. (3 marks)

Show Answers Only

`1/2 ln 2 \ \ text(or)\ \ ln\ sqrt2`

Show Worked Solution

`int_0^(pi/4) tan\ x\ dx`	`=int_0^(pi/4) (sin\ x)/(cos\ x)\ dx`
	`=[-ln\ cos\ x]_0^(pi/4)`
	`=[-ln\ cos\ pi/4 – (-ln cos 0)]`
	`=-ln\ 1/sqrt2 + ln\ 1`
	`=ln sqrt2`
	`=1/2 ln 2`

Calculus, EXT2 C1 2010 HSC 1a

Find `int x/(sqrt(1 + 3x^2))\ dx`. (2 marks)

Show Answers Only

`1/3 sqrt(1 + 3x^2) + c`

Show Worked Solution

`text(Let)\ u = 1 + 3x^2, \ du = 6x\ dx`

`:.int x/(sqrt(1 + 3x^2))\ dx`	`=1/6 int u^(-1/2)\ du`
	`=1/6 xx 2 xx sqrtu + c`
	`=1/3 sqrt(1 + 3x^2) + c`

Calculus, EXT2 C1 2011 HSC 7b

Let `I = int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx.`

Use the substitution `u = 4-x` to show that

`I = int_1^3 (sin^2(pi/8 u))/(u(4-u))\ du.` (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, find the value of `I`. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. `text(Proof)\ \ text{(See Worked Solutions)}`

b. `1/4 log_e 3`

Show Worked Solution

a. `text(Let)\ \ u = 4-x,\ \ du = -dx,\ \ x = 4-u`

`text(When)\ \ x = 1,\ \ u = 3`

`text(When)\ \ x = 3,\ \ u = 1`

`I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx`
	`=int_3^1 (cos^2 (pi/8(4-u)))/((4-u)u) (-du)`
	`=-int_3^1 (cos^2(pi/2-pi/8 u))/((4-u)u)\ du`
	`=int_1^3 (sin^2(pi/8 u))/(u (4-u))\ du`

b. `text{S}text{ince}\ \ int_1^3 (cos^2(pi/8 x))/(x(4-x))\ dx = int_1^3 (sin^2(pi/8 x))/(x(4-x))\ dx`

`text(We can add the integrals such that)`

`2I`	`=int_1^3 (cos^2(pi/8 x))/(x(4-x)) dx + int_1^3 (sin^2(pi/8 x))/(x(4-x)) dx`
	`=int_1^3 1/(x(4-x)) dx`

`text(Using partial fractions:)`

♦ Mean mark 36%.

`1/(x(4-x))`	`=A/x+B/(4-x)`
`1`	`=A(4-x)+Bx`

`text(When)\ \ x=0,\ \ A=1/4`

`text(When)\ \ x=0,\ \ B=1/4`

`2I`	`=1/4 int_1^3 (1/x + 1/(4-x))\ dx`
	`=1/4 [log_e x-log_e (4-x)]_1^3`
	`=1/4 [log_e 3-log_e 1-(log_e 1-log_e 3)]`
	`=1/2 log_e 3`
`:.I`	`=1/4 log_e 3`

Calculus, EXT2 C1 2011 HSC 1b

Evaluate `int_0^3 x sqrt (x + 1)\ dx.` (3 marks)

--- 7 WORK AREA LINES (style=lined) ---

Show Answers Only

`116/15`

Show Worked Solution

`text(Let)\ \ u = 1 + x,\ \ \ du = dx`

`text(When)\ \ x = 0,`	`\ \ \ u = 1`
`x = 3,`	`\ \ \ u = 4`

`int_0^3 x sqrt (1 + x)\ dx`	`= int_1^4 (u – 1) sqrt u\ du`
	`= int_1^4 (u^(3/2) – u^(1/2))\ du`
	`= [2/5 u^(5/2) – 2/3 u^(3/2)]_1^4`
	`= (2/5 xx 32 – 2/3 xx 8) – (2/5 – 2/3)`
	`= 116/15`

Calculus, EXT2 C1 2012 HSC 11e

Evaluate `int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`. (3 marks)

Show Answers Only

`1/2log_e((e^2 + 1)/2)`

Show Worked Solution

`int_0^1 (e^(2x))/(e^(2x) + 1)\ dx`	`= 1/2[log_e(e^(2x) + 1)]_0^1`
	`= 1/2[log_e(e^2 + 1) − log_e2]`
	`= 1/2log_e((e^2 + 1)/2)`

Calculus, EXT2 C1 2013 HSC 14a

The diagram shows the graph `y = ln x.`

By comparing relevant areas in the diagram, or otherwise, show that

`ln t > 2 ((t - 1)/(t + 1))`, for `t > 1.` (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`text(Area under curve)`	`>\ text(Area of triangle)`
`int_1^t ln x\ dx`	`> 1/2 xx (t – 1)ln t,\ \ \ t > 1`
`underbrace{int_1^t 1\ln x\ dx}_text(integration by parts)`	`> 1/2 xx (t – 1)ln t`
`[x ln x]_1^t – int_1^t x * 1/x\ dx`	`> ((t – 1)ln t)/2`
`(tlnt – ln 1) – [x]_1^t`	`> ((t – 1)ln t)/2`
`t ln t – (t – 1)`	`> ((t – 1) ln t)/2`
`2t ln t – 2t + 2`	`> t ln t – ln t`
`t ln t + ln t`	`> 2(t – 1)`
`(t + 1) ln t`	`> 2 (t – 1)`
`ln t`	`> 2 ((t – 1)/(t + 1)),\ \ \ t > 1`

Calculus, EXT2 C1 2013 HSC 11d

Evaluate `int_0^1 x^3 sqrt(1 - x^2)\ dx.` (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

`2/15`

Show Worked Solution

STRATEGY: The choice of `u=1-x^2` or `u^2=1-x^2` provided much less calculation than `u=x^2`. Take note!

`text(Let)\ \ u = 1 – x^2\ ,\ \ du = -2x\ dx`

`text(When)\ \ x = 0\ , \ \ u = 1`

`text(When)\ \ x = 1\ , \ \ u = 0`

`int_0^1 x^3 sqrt(1 – x^2)\ dx`	`= int_0^1 x^2 sqrt (1 – x^2) \ x\ dx`
	`= int_1^0 (1 – u) sqrt u xx ((-du)/2)`
	`= 1/2 int_0^1 (u^(1/2) – u^(3/2))\du`
	`= 1/2 [2/3 u^(3/2) – 2/5 u^(5/2)]_1^0`
	`= 1/2 [(2/3 – 2/5)-0]`
	`= 2/15`