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

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6. (a) Find $ \int \tan x \, dx $ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 3

Question 1

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

Worked Solution & Example Answer:6. (a) Find $ \int \tan x \, dx $ - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 3

Step 1

Find $ \int \tan x \, dx $

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Answer

To find the integral of ( \tan x ), we can use the identity ( \tan x = \frac{\sin x}{\cos x} ). Thus, we have:

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$

This can be solved using substitution: let ( u = \cos x ), which gives ( du = -\sin x , dx ). Therefore, the integral becomes:

$-\int \frac{1}{u} \, du = -\ln|u| + c = -\ln|\cos x| + c$

Consequently, the result is:

$\int \tan x \, dx = -\ln|\cos x| + c$

Step 2

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

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Answer

To solve this using integration by parts, we let:

( u = \ln x )
( dv = \frac{1}{x^3} , dx )

We differentiate and integrate:

( du = \frac{1}{x} , dx )
( v = -\frac{1}{2x^2} )

Applying the integration by parts formula ( \int u , dv = uv - \int v , du ):

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

Which simplifies to:

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

Now integrating ( x^{-3} ) gives:

$= -\frac{\ln x}{2x^2} - \frac{1}{4x^2} + c$

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^{-x} \ln(1 + e^x) + k $

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Answer

Using the substitution ( u = 1 + e^x ), we find:

Differentiate to get ( du = e^x , dx ) or ( dx = \frac{du}{e^x} = \frac{du}{u - 1} ).
Substitute into the integral:

$\int \frac{e^x}{1 + e^x} \, dx = \int \frac{1}{u} \, du$

Integrating gives:

$= \ln|u| + c = \ln(1 + e^x) + c$

Now, to find ( e^{-x} ), we can express it as follows:

$= \frac{1}{e^x} = \frac{1}{u - 1}$

By back-substituting:

Therefore, we can derive:

$\int \frac{e^x}{1 + e^x} \, dx = \frac{1}{2} e^{-x} - e^{-x} \ln(1 + e^x) + k$

6. (a) Find \( \int \tan x \, dx \) - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 3

Worked Solution & Example Answer:6. (a) Find \( \int \tan x \, dx \) - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 3

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

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

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

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