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

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

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 3 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

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 2 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

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 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 = 𝟏 𝟒 , 𝒂= 𝟏 𝟐 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 − 1 8 𝑑𝑢=𝑥 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

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 2 +9 +𝐵 arctan 𝑥 3 +𝐶 𝒂 𝟐 =𝟗, 𝒂=𝟑 𝑥+2 𝑥 2 +9 𝑑𝑥= 𝑥 𝑥 2 +9 𝑑𝑥+ 2 𝑥 2 +9 𝑑𝑥 a) 𝑢= 𝑥 2 +9 1 2 𝑑𝑢=𝑥 𝑑𝑥 𝑥 𝑥 2 +9 𝑑𝑥= 1 2 1 𝑢 𝑑𝑢 = 1 2 ln 𝑢 +𝐶 = 1 2 ln 𝑥 2 +9 + 𝐶 1 2 𝑥 2 +9 𝑑𝑥 =2× 1 3 arctan 𝑥 3 + 𝐶 2 Question from 2018 assessment materials 𝑥+2 𝑥 2 +9 𝑑𝑥= 1 2 ln 𝑥 2 +9 + 𝐶 1 + 2 3 arctan 𝑥 3 + 𝐶 2 = 1 2 ln 𝑥 2 +9 + 2 3 arctan 𝑥 3 +𝐶

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 1 6 ln 2 + 1 18 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 = 1 2 ln 𝑥 2 +9 + 2 3 arctan 𝑥 3 3 0 b) = 1 2 ln 18 + 2 3 arctan 1 − 1 2 ln 9 − 2 3 arctan 0 = 1 2 ln 2 + 𝜋 6 mean value of f(x) = 1 b−a 0 3 𝑓 𝑥 𝑑𝑥 = 1 3 1 2 ln 2+ 𝜋 6 = 1 6 ln 2 + 1 18 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 2 + 1 18 𝜋+𝐥𝐧 𝐤 = 1 6 ln 2 + 1 6 ln 𝑘 6 + 1 18 𝜋 = 1 6 ln 2 𝑘 6 + 1 18 𝜋

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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