1. Methods in calculus

2. FM Methods in Calculus: inverse trig functions
KUS objectives
BAT differentiate and integrate inverse trig functions using techniques from A2
Starter: find 𝒅𝒚 𝒅𝒙 in terms of x and y for the following
𝑑𝑦 𝑑𝑥 =− 𝑥 𝑦
𝑥 2 + 𝑦 2 =1
5 𝑥 2 +𝑥𝑦+2 𝑦 2 =11
𝑑𝑦 𝑑𝑥 = 10𝑥+𝑦 𝑥+4𝑦
𝑥= tan 𝑦
𝑑𝑦 𝑑𝑥 = 1 𝑠𝑒𝑐 2 𝑦

3. cos y× dy dx = 1 = 1 𝑐𝑜𝑠 2 𝑦 = 1 1− 𝑠𝑖𝑛 2 𝑦 = 1 1− 𝑥 2 dy dx = 1 cos 𝑦
WB C1 a) show that 𝑑 𝑑𝑥 arcsin 𝑥 = − 𝑥 2
Let y= arcsin 𝑥
Then sin y=sin arcsin 𝑥 =𝑥
Differentiate implicitly
cos y× dy dx = 1
Rearrange and use identity
= 𝑐𝑜𝑠 2 𝑦
= − 𝑠𝑖𝑛 2 𝑦
= − 𝑥 2
dy dx = 1 cos 𝑦
Now b) show that 𝑑 𝑑𝑥 arccos 𝑥 =− − 𝑥 2
and c) show that 𝑑 𝑑𝑥 arc𝑡𝑎𝑛 𝑥 = 𝑥 2

4. −sin y× dy dx = 1 =− 1 𝑠𝑖𝑛 2 𝑦 =− 1 1− 𝑐𝑜𝑠 2 𝑦 =− 1 1− 𝑥 2
C1 (cont) b) show that 𝑑 𝑑𝑥 arccos 𝑥 =− − 𝑥 2
Let y= arccos 𝑥
Differentiate implicitly
Then cos y=𝑥
−sin y× dy dx = 1
Rearrange and use identity
=− 1 𝑠𝑖𝑛 2 𝑦
=− 1 1− 𝑐𝑜𝑠 2 𝑦
=− 1 1− 𝑥 2
dy dx =− 1 sin 𝑦
C c) show that 𝑑 𝑑𝑥 arc𝑡𝑎𝑛 𝑥 = 𝑥 2
Let y= arctan 𝑥
Then tan y=𝑥
Differentiate implicitly
𝑠𝑒𝑐 2 𝑦× dy dx = 1
Rearrange and use identity 𝑡𝑎𝑛 2 𝑦+1= 𝑠𝑒𝑐 2 𝑦
dy dx = 1 𝑠𝑒𝑐 2 𝑦
= 1 𝑡𝑎𝑛 2 𝑦+1
= 1 1+ 𝑥 2

5. Notes Implicit differentiation
For some equations it is impossible to rearrange to give y = f(x) and hence differentiate. One approach is to use the chain rule to differentiate each term without rearrangement
For example differentiate y2
Think
Pair
share
Now we can do
Not possible to separate the variables – do by inspection

6. One approach is to use the chain rule to differentiate each term without rearrangement
Key examples:
𝑑 𝑑𝑥 𝑦 𝑛 =𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
function
Gradient function
𝒚 𝟒 =𝟑𝒙
𝟏 𝒚 = 𝟏 𝒙
𝒚 =𝒄𝒐𝒔 𝒙
𝐬𝐢𝐧 𝒚 =𝟑 𝒙 𝟐
𝒙 𝟐 + 𝒚 𝟐 =𝟏𝟔
𝒆 𝒚 = 𝒆 −𝟑𝒙
𝐥𝐧 𝒚 =𝒍𝒏 𝒙+𝟓𝒙
4 𝑦 3 𝑑𝑦 𝑑𝑥 =3
→ 𝑑𝑦 𝑑𝑥 = 3 4 𝑦 if needed
− 1 𝑦 2 𝑑𝑦 𝑑𝑥 =− 1 𝑥 2
1 2 𝑦 𝑑𝑦 𝑑𝑥 =−sin x
cos 𝑦 𝑑𝑦 𝑑𝑥 =6𝑥
2𝑥+2𝑦 𝑑𝑦 𝑑𝑥 =0
𝑒 𝑦 𝑑𝑦 𝑑𝑥 = −3𝑒 −3𝑥
1 𝑦 𝑑𝑦 𝑑𝑥 = 1 𝑥 +5

7. cos y× dy dx = 2𝑥 = 2𝑥 1− 𝑠𝑖𝑛 2 𝑦 = 2𝑥 1− 𝑥 4 dy dx = 2𝑥 cos 𝑦
WB C2 Given 𝑦= arcsin 𝑥 2 , find 𝑑𝑦 𝑑𝑥
a) Using implicit differentiation
Using the chain rule and the formula for 𝑑 𝑑𝑥 arcsin 𝑥
a) Let y= arcsin 𝑥 2
Then sin y=sin arcsin 𝑥 2 = 𝑥 2
Differentiate implicitly
cos y× dy dx = 2𝑥
Rearrange and use identity
= 2𝑥 1− 𝑠𝑖𝑛 2 𝑦
= 2𝑥 1− 𝑥 4
dy dx = 2𝑥 cos 𝑦
b) Let t= 𝑥 2 then y= arcsin 𝑡 Differentiate parametric equations
Then dt dx =2𝑥 and dy dt = 1 1− 𝑡 2
use chain rule
So dy dx = dt dx × dy dt =2𝑥× − 𝑡 2
So dy dx = 2𝑥 1− 𝑥 4

8. NOW DO EX 3C 𝑠𝑒𝑐 2 𝑥 d𝑦 𝑑𝑥 = −2 (1+𝑥) 2 dy dx = 1 𝑠𝑒𝑐 2 𝑥 × −2 (1+𝑥) 2
WB C3 Given 𝑦= 1−𝑥 1+𝑥 find 𝑑𝑦 𝑑𝑥
Let tan y= 1−𝑥 1+𝑥 Differentiate RHS implicitly and LHS by quotient rule
𝑠𝑒𝑐 2 𝑥 d𝑦 𝑑𝑥 = 1+𝑥 −1 −(1−𝑥)(1) (1+𝑥) 2
𝑠𝑒𝑐 2 𝑥 d𝑦 𝑑𝑥 = −2 (1+𝑥) 2
Rearrange and use identity
dy dx = 1 𝑠𝑒𝑐 2 𝑥 × −2 (1+𝑥) 2
= 𝑡𝑎𝑛 2 𝑥 × −2 (1+𝑥) 2
dy dx = −𝑥 1+𝑥 2 × −2 (1+𝑥) 2
=…= (1+𝑥) 𝑥 2 × −2 (1+𝑥) 2
𝑑𝑦 𝑑𝑥 = −1 2+2 𝑥 2
NOW DO EX 3C

9. WB D1 By using an appropriate substitution, show that 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶 where a is a positive constant and 𝑥 <𝑎
1 𝑎 2 − 𝑥 2 𝑑𝑥= 𝑎 2 1−− 𝑥 𝑎 𝑑𝑥=
= 1 𝑎 − 𝑥 𝑎 𝑑𝑥
𝑢= 𝑎 𝑥
𝑑𝑢= 1 𝑎 𝑑𝑥
= 1 𝑎 − 𝑢 𝑑𝑥
= arcsin 𝑢 +𝐶
= arcsin 𝑥 𝑎 +𝐶

10. WB D1 (cont) By using an appropriate substitution, show that a) 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶 where a is a positive constant and 𝑥 <𝑎
1 𝑎 2 − 𝑥 2 𝑑𝑥= 𝑎 2 1−− 𝑥 𝑎 𝑑𝑥=
= 1 𝑎 − 𝑥 𝑎 𝑑𝑥
𝑢= 𝑥 𝑎
𝑑𝑢= 1 𝑎 𝑑𝑥
= − 𝑢 𝑑𝑢
= arcsin 𝑢 +𝐶
= arcsin 𝑥 𝑎 +𝐶
Now show that b) 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶

11. WB D1 (cont) By using an appropriate substitution, show that b) 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶 where a is a positive constant and 𝑥 <𝑎
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 𝑥 𝑎 𝑑𝑥=
𝑢= 𝑥 𝑎
𝑑𝑢= 1 𝑎 𝑑𝑥
= 1 𝑎 𝑢 2 𝑑𝑢
= 1 𝑎 arctan 𝑢 +𝐶
= 1 𝑎 arctan 𝑥 𝑎 +𝐶

12. WB D2 Find 4 5+ 𝑥 2 𝑑𝑥 4 5+ 𝑥 2 𝑑𝑥= 4 1 5+ 𝑥 2 𝑑𝑥 𝑎 2 =5
1 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶
4 5+ 𝑥 2 𝑑𝑥= 𝑥 2 𝑑𝑥
𝑎 2 =5
= arctan 𝑥 𝐶
= arctan 𝑥 𝐶

13. WB D3 Find 𝑥 2 𝑑𝑥
1 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶
𝑎 2 = 25 9
𝑥 2 𝑑𝑥= 𝑥 2 𝑑𝑥
= arctan 𝑥 𝐶
= arctan 3𝑥 5 +𝐶

14. WB D4 Find − 3 /4 3 /4 1 3−4 𝑥 2 𝑑𝑥 𝑎 2 = 3 4 − 3 /4 3 /4 1 3−4 𝑥 2 𝑑𝑥
1 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶
𝑎 2 = 3 4
− 3 /4 3 / −4 𝑥 2 𝑑𝑥
= − 3 /4 3 / − 𝑥 2 𝑑𝑥
= arcsin 𝑥 /4 − 3 /4
= arcsin − arcsin − 1 2
= 𝜋 6

15. 𝑥+4 1−4 𝑥 2 𝑑𝑥=− 1 4 1−4 𝑥 2 +2 arcsin 2𝑥 +C
WB D5 Find 𝑥+4 1−4 𝑥 𝑑𝑥
1 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶
𝒂 𝟐 = 𝟏 𝟒 , 𝒂= 𝟏 𝟐
𝑥+4 1−4 𝑥 2 = 𝑥 1−4 𝑥 −4 𝑥 2
𝑥 1−4 𝑥 2 𝑑𝑥 =− 𝑢 𝑑𝑢
𝑢=1−4 𝑥 2
− 1 8 𝑑𝑢=𝑥 𝑑𝑥
=− 1 4 𝑢 𝐶 = − −4 𝑥 2 +𝐶
−4 𝑥 2 𝑑𝑥
= − 𝑥 2 𝑑𝑥
=2 arcsin 2𝑥 +𝐶
𝑥+4 1−4 𝑥 𝑑𝑥=− −4 𝑥 arcsin 2𝑥 +C

16. Show that 𝑓(𝑥) 𝑑𝑥=𝐴 ln 𝑥 2 +9 +𝐵 arctan 𝑥 3 +𝐶
1 𝑎 2 − 𝑥 2 𝑑𝑥= arcsin 𝑥 𝑎 +𝐶
1 𝑎 2 + 𝑥 2 𝑑𝑥= 1 𝑎 arctan 𝑥 𝑎 +𝐶
WB D6a 𝑓 𝑥 = 𝑥+2 𝑥 2 +9
Show that 𝑓(𝑥) 𝑑𝑥=𝐴 ln 𝑥 𝐵 arctan 𝑥 3 +𝐶
𝒂 𝟐 =𝟗, 𝒂=𝟑
𝑥+2 𝑥 𝑑𝑥= 𝑥 𝑥 𝑑𝑥 𝑥 𝑑𝑥 a)
𝑢= 𝑥 2 +9
1 2 𝑑𝑢=𝑥 𝑑𝑥
𝑥 𝑥 𝑑𝑥= 𝑢 𝑑𝑢
= 1 2 ln 𝑢 +𝐶 = ln 𝑥 𝐶 1
2 𝑥 𝑑𝑥
=2× 1 3 arctan 𝑥 𝐶 2
Question from 2018 assessment materials
𝑥+2 𝑥 𝑑𝑥= 1 2 ln 𝑥 𝐶 arctan 𝑥 𝐶 2
= ln 𝑥 arctan 𝑥 3 +𝐶

17. NOW DO EX 3D 0 3 𝑓 𝑥 𝑑𝑥 = 1 2 ln 𝑥 2 +9 + 2 3 arctan 𝑥 3 3 0 b)
WB D6bc (cont) 𝑓 𝑥 = 𝑥+2 𝑥 2 +9
b) Show that the mean value of f(x) over the interval 0, 3 is ln 𝜋
c) Use (a) to find the mean value over the interval 0, 3 of 𝑓 𝑥 + ln 𝑘
Where k is a positive constant, giving your answer in the form p+ 1 6 ln q
0 3 𝑓 𝑥 𝑑𝑥 = ln 𝑥 arctan 𝑥 b)
= ln arctan 1 − ln 9 − 2 3 arctan 0
= ln 2 + 𝜋 6
mean value of f(x) = 1 b−a 0 3 𝑓 𝑥 𝑑𝑥
= ln 2+ 𝜋 6 = 1 6 ln 𝜋 QED
𝑐) 𝑓 𝑥 + ln 𝑘 is a shift in the y-direction of + ln 𝑘
so the mean value over the interval will be the area
Question from 2018 assessment materials
1 6 ln 𝜋+𝐥𝐧 𝐤
= 1 6 ln ln 𝑘 𝜋
= 1 6 ln 2 𝑘 𝜋

18. One thing to improve is –
KUS objectives
BAT differentiate and integrate inverse trig functions using techniques from A2
self-assess
One thing learned is –
One thing to improve is –

19. END