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. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. 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

13. NOW DO EX 3D WB D5 Find 𝑥+4 1−4 𝑥 2 𝑑𝑥 𝑎 2 = 1 4
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
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
𝑑𝑢=8𝑥 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
NOW DO EX 3D

14. 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 –

15. END