9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx. \] It is given that, for $n \geq 0$, \[ I_n = \int_0^{\pi/4} \sec^2 x \tan^2 x \, dx. \] (ii) Using the result that $ \frac{d}{dx} (\sec x) = \tan x \sec x $, show that, for $n \geq 2$, \[ (n + 1)I_n = (\sqrt{2})^n + (n - 2)I_{n-2}. \]

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9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx - CIE - A-Level Further Maths - Question 9 - 2018 - Paper 1

Question 9

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9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx. \] It is given that, for $n \geq 0$, \[ I_n = \int_0^{\pi/4} \sec^2... show full transcript

Worked Solution & Example Answer:9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx - CIE - A-Level Further Maths - Question 9 - 2018 - Paper 1

Step 1

Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx. \]

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Answer

To solve [ \int \sec^2 x \tan^2 x , dx, ] we use the substitution $u = \tan x$ . Thus, we have:

Calculate $du = \sec^2 x \, dx$ .
The integral becomes: [ \int u^2 , du = \frac{u^3}{3} + C = \frac{\tan^3 x}{3} + C. ]

Step 2

Using the result that $ \frac{d}{dx} (\sec x) = \tan x \sec x $, show that, for $n \geq 2$, \[ (n + 1)I_n = (\sqrt{2})^n + (n - 2)I_{n-2}. \]

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Answer

To show this statement, start with: [ I_n = \int_0^{\pi/4} \sec^2 x \tan^2 x , dx. ]

Using integration by parts:

Let ( u = \tan x ) and ( dv = \sec^2 x , dx ).
Then, ( du = \sec^2 x , dx ) and ( v = \tan x ).
This leads to: [ I_n = \tan x \tan^2 x \bigg|_0^{\pi/4} - \int_0^{\pi/4} \tan x \cdot 2 \tan x \sec^2 x , dx. ]
Finally, substituting and simplifying, we find: [ (n + 1)I_n = (\sqrt{2})^n + (n - 2)I_{n-2}. ]

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9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx - CIE - A-Level Further Maths - Question 9 - 2018 - Paper 1

Worked Solution & Example Answer:9 (i) Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx - CIE - A-Level Further Maths - Question 9 - 2018 - Paper 1

Using the substitution $u = \tan x$, or otherwise, find \[ \int \sec^2 x \tan^2 x \, dx. \]

Using the result that \( \frac{d}{dx} (\sec x) = \tan x \sec x \), show that, for $n \geq 2$, \[ (n + 1)I_n = (\sqrt{2})^n + (n - 2)I_{n-2}. \]

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