Methods in calculus

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

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

−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 =− 1 1− 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 C1 c) show that 𝑑 𝑑𝑥 arc𝑡𝑎𝑛 𝑥 = 1 1+ 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

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× 1 1− 𝑡 2 So dy dx = 2𝑥 1− 𝑥 4

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 = 1 1+ 𝑡𝑎𝑛 2 𝑥 × −2 (1+𝑥) 2 dy dx = 1 1+ 1−𝑥 1+𝑥 2 × −2 (1+𝑥) 2 =…= (1+𝑥) 2 2+2 𝑥 2 × −2 (1+𝑥) 2 𝑑𝑦 𝑑𝑥 = −1 2+2 𝑥 2 NOW DO EX 3C

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

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

WB D1 (cont) By using an appropriate substitution, show that b) 1 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 1 1+ 𝑥 𝑎 2 𝑑𝑥= 𝑢= 𝑥 𝑎 𝑑𝑢= 1 𝑎 𝑑𝑥 = 1 𝑎 1 1+ 𝑢 2 𝑑𝑢 = 1 𝑎 arctan 𝑢 +𝐶 = 1 𝑎 arctan 𝑥 𝑎 +𝐶

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= 4 1 5+ 𝑥 2 𝑑𝑥 𝑎 2 =5 =4 1 5 arctan 𝑥 5 +𝐶 = 4 5 arctan 𝑥 5 +𝐶

WB D3 Find 1 25+9 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 1 25+9 𝑥 2 𝑑𝑥= 1 9 1 5 3 2 + 𝑥 2 𝑑𝑥 = 1 9 1 5 3 arctan 𝑥 5 3 +𝐶 = 1 15 arctan 3𝑥 5 +𝐶

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 1 3−4 𝑥 2 𝑑𝑥 = 1 2 − 3 /4 3 /4 1 3 4 − 𝑥 2 𝑑𝑥 = 1 2 arcsin 𝑥 3 2 3 /4 − 3 /4 = 1 2 arcsin 1 2 − arcsin − 1 2 = 𝜋 6

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 2 + 4 1−4 𝑥 2 𝑥 1−4 𝑥 2 𝑑𝑥 =− 1 8 1 𝑢 𝑑𝑢 𝑢=1−4 𝑥 2 𝑑𝑢=8𝑥 𝑑𝑥 =− 1 4 𝑢 1 2 +𝐶 = − 1 4 1−4 𝑥 2 +𝐶 4 1 1−4 𝑥 2 𝑑𝑥 =2 1 1 4 − 𝑥 2 𝑑𝑥 =2 arcsin 2𝑥 +𝐶 𝑥+4 1−4 𝑥 2 𝑑𝑥=− 1 4 1−4 𝑥 2 +2 arcsin 2𝑥 +C NOW DO EX 3D

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 –

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