6. (a) Find $ \int \tan^3 x \, dx $. (b) Use integration by parts to find $ \int \frac{1}{x^3} \ln x \, dx $. (c) Use the substitution $ u = 1 + e^x $ to show that $ \int \frac{e^x}{1+e^x} \, dx = \frac{1}{2} e^{-x} - e^{-2x} + \ln(1+e^x) + k, $ where $ k $ is a constant.

A-Level Edexcel Maths Pure Applications of Differentiation: 6. (a) Find \( \int \tan^3 x \,

6. (a) Find $ \int \tan^3 x \, dx $ - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 3

Question 1

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6. (a) Find $ \int \tan^3 x \, dx $. (b) Use integration by parts to find $ \int \frac{1}{x^3} \ln x \, dx $. (c) Use the substitution $ u = 1 + e^x $ to sh... show full transcript

Worked Solution & Example Answer:6. (a) Find $ \int \tan^3 x \, dx $ - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 3

Step 1

Find $ \int \tan^3 x \, dx $

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Answer

To find ( \int \tan^3 x , dx ), we use the identity ( \tan^2 x = \sec^2 x - 1 ):

\int \tan^3 x \, dx = \int \tan x (\sec^2 x - 1) \, dx = \int \tan x \sec^2 x \, dx - \int \tan x \, dx.

Let ( u = \tan x ). Then ( du = \sec^2 x , dx ) and:

\int \tan^3 x \, dx = \int u \, du - \int u \, dx = \frac{u^2}{2} - \ln |\sec x + \tan x| + C = \frac{\tan^2 x}{2} - \ln |\sec x + \tan x| + C.

Step 2

Use integration by parts to find $ \int \frac{1}{x^3} \ln x \, dx $

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Answer

Using integration by parts, we let:

( u = \ln x ) ( \Rightarrow du = \frac{1}{x} , dx )
( dv = \frac{1}{x^3} , dx ) ( \Rightarrow v = -\frac{1}{2x^2} )

Then, applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we have:

\int \frac{1}{x^3} \ln x \, dx = -\frac{1}{2x^2} \ln x - \int -\frac{1}{2x^2} \cdot \frac{1}{x} \, dx.

Calculating the remaining integral:

\int \frac{1}{x^3} \ln x \, dx = -\frac{1}{2x^2} \ln x + \frac{1}{4x^2} + C.

Step 3

Use the substitution $ u = 1 + e^x $ to show that $ \int \frac{e^x}{1+e^x} \, dx = \frac{1}{2} e^{-x} - e^{-2x} + \ln(1+e^x) + k, $

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Answer

To solve ( \int \frac{e^x}{1+e^x} , dx ) using the substitution ( u = 1 + e^x ), we find:

Calculate ( du = e^x , dx ), so that ( dx = \frac{du}{e^x} = \frac{du}{u - 1}

Then, the integral becomes:

\int \frac{e^x}{u} \frac{du}{u-1} = \int \frac{1}{u} \, du = \ln |u| + C = \ln(1+e^x) + C.

Integrate.

The integral evaluates to:

= \frac{1}{2} e^{-x} - e^{-2x} + \ln(1+e^x) + k, $$\nwhere \( k \) is a constant.

6. (a) Find \( \int \tan^3 x \, dx \) - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 3

Worked Solution & Example Answer:6. (a) Find \( \int \tan^3 x \, dx \) - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 3

Find \( \int \tan^3 x \, dx \)

Use integration by parts to find \( \int \frac{1}{x^3} \ln x \, dx \)

Use the substitution \( u = 1 + e^x \) to show that \( \int \frac{e^x}{1+e^x} \, dx = \frac{1}{2} e^{-x} - e^{-2x} + \ln(1+e^x) + k, \)

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